# Rd Sharma Solutions Class 8 Chapter 6 Algebraic Expressions and Identities

Welcome to NCTB Solution. Here with this post we are going to help 8th class students for the Solutions of RD Sharma Class 8 Mathematics, Chapter 6, Algebraic Expressions and Identities. Here students can easily find Exercise wise solution for chapter 6, Algebraic Expressions and Identities. Students will find proper solutions for Exercise 6.1, 6.2, 6.3, 6.4, 6.5, 6.6 and 6.7 Our teacher’s solved every problem with easily understandable methods so that every students can understand easily.

Algebraic Expressions and Identities Exercise 6.1 Solution :

Question no – (1)

Solution :

(i) 7x²yz – 5xy

Terms = 7xyz, – 5xy

Coefficient = 7, -5

(ii) x²+ x+1

Terms = x², x, 1

Coefficient =1, 1, 1

(iii) 3x² y² – 5x²y²z²+z²

Terms = 3x²y², 5z²y²z², z²

Coefficient = 3, 5, 1, 9

(iv) 9 – ab + bc – ca

Terms = -ab, bc, -ca, a/2, b/2, -ab

Coefficient = 1/2, 1/2, -1

(v) a/2 + b/2 –ab

Terms = a/2 , b/2, -ab

Coefficient = 1/2, 1/2, -1

(vi) 0.2x – 0.3xy + 0.5y

Terms = 0.2x, – 0.3xy, 0.5y

Coefficient = 0.2, -0.3, 0.5

Question no – (2)

Solution :

(i) x + y

= Binomial

(ii) 1000

= Monomial

(iii) x + x² + x³ + x⁴

= None of these

(iv) 7+ a + 5b

= Trinomial

(v) 2b – 3b²

= Binomial

(vi) 2y – 3y² + 4y³

= Trinomial

(vii) 5x- 4y+3x

= Trinomial

(viii) 4a – 15a²

= Binomial

(ix) xy + yz + zt + tx

= None of these

(x) pqr

= Monomial

(xi) p²q + pq²

= Binomial

(xii) 2p + 2q

= Binomial

Algebraic Expressions and Identities Exercise 6.2 Solution :

Question no – (1)

Solution :

(i) 3a²b, -4a²b, 9a²b

= 3a²b – 4a²b + 9a²b

= 8a²b

(ii) 2/3a, 3/5a,- 6/5a

= 2/3a + 3/5a – 6/5a

= 10a + 9a – 18a/15

= 19a – 18a/15

= 1a/15

(iii) 4xy² – 7x²y – 12x²y – 6xy² – 3x²y + 5xy²

= 4xy² – 7x²y + 12x²y – 6xy² – 3x²y + 5xy²

= 3xy² + 2x²y

(iv) 3/2a – 5/4b + 2/5c, 2/3a – 7/2b + 7/2c, 5/3a + 5/2b – 5/4c

= 3/2a – 5/4b + 2/5c + 2/3a – 7/2b + 7/2c + 5/3a + 5/2b – 5/4c

= (3/2a + 2/3a + 5/3a) – (5/4b + 7/2b – 5/2b) + (2/5c + 7/2c – 5/4c)

= [(9 + 4 + 10/6)a] + [(5 + 14 – 10/4) b] + [(8 + 70 – 25/20) c]

= 23a/6 – 9b/4 + 53c/20

(vi) (11/2xy + 12/5y + 13/7x) + (- 11/2y – 12/5x – 13/7xy)

= (11/2 – 13/7) xy + (12/5 -11/2)y + (13/7 – 12/5)x

= (77 – 26/14)xy + (24 – 55/10)y + (5 – 84/35)x

= 51/14xy – 31/10y – 19/35x

Question no – (2)

Solution :

(i) -5xy from 12xy

= 12xy – (-5xy)

= 12xy + 5xy

= 17xy

(ii) 2a² from – 7a²

= -7a² – (2a²)

= -7a² – 2a²

= -9a²

(iv) 2x³ – 4x² + 3x+ 5 from 4x³ +x²+ x+6

= (4x³ + x²+ x +6) – (2x³ – 4x² + 3x + 5)

= 4x³ – 2x³ + 4x² + x² + x- 3x + 6 – 5

= 2x³ + 5x² – 2x +1

(v) 2/3y³ -2/7y² – 5 from 1/3y³ + 5/7y² + y – 2

= (1/3y³ + 5/7y² + y – 2) – (2/3y³ – 2/7y² – 5)

= 1/3y³ – 2/3y³ + 5/7y² + y + 3

= 1-2/3y³ + 5/7 + 2y² + y + 3

= -1/3 y³ + 7/7y² + 2/7y² + y + 3

= -1/3y³ + y² + y + 3

(vi) Given, x²y – 4/5xy² + 4/3xy from 2/3x²y + 3/2xy² – 1/3xy

= (2/3x²y + 3/2xy² – 1/3xy) – (x²y – 4/5xy² + 4/3xy)

= (2/3 – 1)x²y + (3/2 + 4/5)xy – (1/3 – 4/3)xy

= (2 – 3/3)x²y + (15 + 8/10)xy² – (1 + 4/3)xy

= -1/3x²y + 23/10xy² – 5/3xy

Question no – (3)

Solution :

(ii) 5a²/2 + 3a³/2 + a/3 – 6/5 from 1/3a³ – 3/4 a² – 5/2

= (1/3a² – 3/4a² – 5/2) – (5a²/2 + 3a²/2 + a/3 – 6/5)

= 1/3a³ – 3/2a³ – 3/4a² – 3/4a² 5/2a² – 5/2 + 6/5 – a/3

= (1/3 – 3/2)a³ – (3/4 + 5/2)a² – (5/2 – 6/5) – a/3

= (2 – 9/6)a³ – (3 + 10/4)a² – (25 -12/10)

= -7/6a³ – 13/4a² – a/3 – 13/10

(iii) 7/4x³ + 3/5x² + 1/2x + 9/2 from 7/2 – x/3 – x²/5

= (7/2 – x/3 – x²/5) – (7/4x³ + 3/5x² + 1/2x + 9/2)

= 7/2 – x/3 – x²/5 – 7/4x³ + 3/5x² + 1/2x + 9/2

= -7/4x³ – (1 + 3/5)x² – (1/3 – 1/2)x + (7/2 – 9/2)

= – 7/4x³ – 4/5x² + 1/6x – 2/2

= – 7/4x³ – 4/5x² + 1/6x – 1

(iv) y³/3 + 7/3y² + 1/2y + 1/2 from 1/3 -5/3y²

= (1/3 – 5/3y²) + (y³/3 + 7/3y² + 1/2y + 1/2)

= 1/3 – 5/3y² – y³/3 + 7/3y² + 1/2y + 1/2 y

= (1/3 – 1/2) – (5/3y² + 7/3y²) – y³/3 + 1/2y

= (2 – 3/6) – (5 + 7/7)b² – y³/3 + 1/2y

= – 1/6 – 12/3 y² – y³/3 + 1/2y

= -4y² – y³/3 – 1/6 + 1/2y

= – 4y² – y³/3 + 1/2y – 1/6

= -y³/3 – 4y² + 1/2 y – 1/6

(v) 2/3ac – 5/7ab +2/3bc from 3/2ab -7/4ac – 5/6bc

= (3/2ab -7/4ac – 5/6bc) – (2/3ac – 5/7ab +2/3bc)

= 3/2ab – 7/4ac – 5/6bc – 2/3ac + 5/7ab – 2/3bc

= 3/2ab – 5/7ab – 7/4ac – 2/3ac -5/6bc – 2/3bc

= (3/2 + 5/7)ab – (7/4 + 2/3)ac – (5/6 + 2/3)bc

= (21 + 10/14)ab – (21 + 8/12)ac – (15+12/18)bc

= 31/14ab – 29/12ac – 27/18bc

= 31/14ab – 29/12ac – 3/2bc ….. (after cutting)

Question no – (4)

Solution :

Sum = (x – 3y + 3y + 2z) + (-4x + 9y – 11z)

= x – 3y + 2z – 4x + 9y -11z

= -3x + 6y – 9z

∴ The required number,

= (-3x + 6y – 9z) – (3x – 4y – 7z)

= – 3x + 6y – 9z – 4x + 4y + 7z

= – 6x + 10y – 2z

Question no – (5)

Solution :

Sum₁, = (3l – 4m – 7n²) + (2l + 3m – 4n²)

= 3l – 4m – 7n² + 2l + 3m – 4n²

= 5l – m – 11n²

Sum₂, = (9l + 2m – 3n²) + (- 3l + m + 4n²)

= 9l + 2m – 3n² – 3l + m + 4n²

= 6l + 3m + n²

∴ The required number,

= Sum₂ – Sum₁

= (6l + 3m + n²) – (5l – m – 11n²)

= 6l + 3m + n² – 5l – m – 11n²

= l + 4m + 12n²

Question no – (6)

Solution :

Sum = (2x – x² + 5) + (-4x – 3 + 7x²)

= 2x- x² + 5 – 4x – 3 + 7x²

= – 2x – 6x² + 2

∴ The required number,

= 5 – (- 2x + 6x² + 2)

= 5 + 2x – 6x² – 2

= 2x – 6x² + 3

Question no – (7)

Solution :

(i) x²- 3x + 5 – 1/2(3x² – 5x + 7)

= x²- 3x + 5 – 3/2x² + 5/2x – 7/2

= (1 – 3/2)x² – (3 – 5/2) + (5 – 7/2)

= (2 – 3/2)x² – (6 – 5/2)x + (10 – 7/2)

= -1/2x² – 1/2x + 3/2…(Simplified)

(ii) [5 – 3x + 2y – (2x – y)] – (3x – 7y + 9)

= 5 – 3x + 2y – 2x + y – 3x + 7y – 9

= – 8x + 10y – 4…(Simplified)

(iv) (1/3y² – 4/7y + 11) – (1/7y – 3 + 2y²) – (2/7y – 2/3y² + 2)

= 1/3y² – 4/7y + 11 – 1/7y – 3 + 2y² – 2/7y – 2/3y² – 2

= (1/3 – 2 + 2/3)y² – (4/7 + 1/7 + 2/7)y + (11 + 3 – 2)

= (1 – 6 + 2/3)y² – (4 + 1 + 2/7)y + 12

= -3/3y² – 7/7y + 12

= -y² – y + 12…(Simplified)

Algebraic Expressions and Identities Exercise 6.3 Solution :

Question no – (1)

Solution :

Given, 5x² × 4x³

= (5 × 4) (x² × x³)

= 20x⁵

Hence, the product is 20x⁵

Question no – (2)

Solution :

Given, -3a² × 4b⁴

=(-3 × 4) (a² × b⁴)

= 12a²b⁴

Thus, the product is 12a²b⁴

Question no – (3)

Solution :

Given, (-5xy) × (-3x²yz)

= (-5 × 3) (xy × x²yz)

= -15x³y²z

Therefore, the product is -15x³y²z

Question no – (4)

Solution :

Given, 1/2xy × 2/3x²yz²

= (1/2 × 2/3) (xy × x²yz²)

= 1/3x³y²z²

Thus, the product is 1/3x³y²z²

Question no – (5)

Solution :

Given, (-7/5xy²z) × (13/3x²yz²)

= (-7/5 × 13/3) (xy²z × x²yz²)

= -91/15x³y³z³

Hence, the product is -91/15x³y³z³

Question no – (7)

Solution :

Given, (-1/27a²b²) × (9/2 a³b²c²)

= (- 1/27 × 9/2) (a²b² ×a³b²c²)

= -1/6a⁵b⁴c²

Thus, the product is -1/6a⁵b⁴c²

Question no – (8)

Solution :

Given, (-7xy) × (1/4x²yz)

= (-7 × 1/4) (xy × x²yz)

= -7/4x³y²z

Therefore, the product is -1/6a⁵b⁴c²

Question no – (9)

Solution :

Given, (7ab) × (-5ab²c) × (6abc²)

= – (7 × 5 × 6) (ab × ab²c × abc²)

= – 210a³b⁴c³

Hence, the product is -210a³b⁴c³

Question no – (11)

Solution :

Given, (-4x²) × (-6xy²) × (-3yz²)

= – (4 × 6 × 3) (x² × xy² × z²)

= -72x³y²z²

Thus, the product is -72x³y²z²

Question no – (13)

Solution :

Given, (7/9ab²) × (15/7ac²b) × (- 3/5a²c)

= -(7/9 × 15/7 × 3/5) (ab² × ac²b × a²c)

= -a⁴b³c³

So, the product is -a⁴b³c³

Question no – (16)

Solution :

Given, (4/3 pq²) × (-1/4p²r) × (16p² q² r²)

= – (4/3 × 1/4 ×16) (pq² × p²r × p²q²r²)

= – 16/3p⁵q⁴r³

Hence, the product is – 16/3p⁵q⁴r³

Question no – (17)

Solution :

Given, (2.3xy) × ( 0.1x) × (0.16)

= (2.3 × 0.1 × 0.16) (xy × x)

= 0.0368x²y

Thus, the product is 0.0368x²y

Question no – (18)

Solution :

Given, (3x) × (4x) × (-5x)

= – 60x³

Question no – (19)

Solution :

Given, (4x²) × (- 3x)× (4/5x³)

= – (4 ×3 × 4/5) (x² × x × x³)

= 16/5x⁶

Question no – (20)

Solution :

Given, (5x⁴) × (x²)³ × (2x²)

= (5 × 2 × 2) (x⁴ × x⁶ × x²)

= 20x^12

Question no – (21)

Solution :

Given, (x²)³ × (2x) × (-4x) × (5)

= – (2 × 4 × 5) (x⁶ × x × x)

= 40x⁸

Question no – (22)

Solution :

= (-8x²y⁶) × (-20xy)

= (8 × 20) (x² × y⁶ × xy)

= 160x³y⁷

= 160 (2.5)³ (1)⁷ [∵ x = 2.5, y = 1]

= 160 × 15.625 × 1

= 2500 ….(Verified)

Question no – (23)

Solution :

Given, (3.2x⁶y³) × (2.1x²y²)

= (3.2 × 2.1) (x⁶y³ × x²y²)

= 6.72x⁸y⁵

= 6.72 × (1)⁸ × (0.5)⁵ [∵ x =1 y = 0.5]

= 6.72 × 1 × 0.03125

= 0.21

Question no – (24)

Solution :

(5x⁶) × (- 1.5x²y³) × (- 21xy²)

= (5 × 1.5 × 12) (x⁶ × x²y³ × xy)

= 90x⁹y⁵

= 90(1) (0.5)⁵ [∵ x = 1 y = 0.5]

= 90 × 1 × 0.03125

= 2.8125

Question no – (26)

Solution :

(-8 x²y⁶) × (-20xy)

= (8 × 20) (x²y⁶ × xy)

= 160x³y⁷

= 160 × (2.5) (1)⁷ …[∵ x = 2.5 y = 1]

= 160 × 15.625 × 1

= 2500

Question no – (27)

Solution :

Given, (xy³) × (yx³) × (xy)

= -x⁵ y⁵

Question no – (28)

Solution :

Given, (1/8x²y⁴) × (1/4x⁴y²) ×(xy) × 5

= (1/8 × 1/4 × 5) (x²y⁴ × x⁴y² × xy)

= 5/32x⁷y⁷

Question no – (29)

Solution :

Given, (2/5a²b) × (-15b²ac) × (-1/2c²)

= (2/8 × 15 × 2/1) (a²b × b²ac × c²)

= 3a³b³c³

Question no – (30)

Solution :

Given, (-4/7a²b) × (-2/3b²c) × (-7/6c²a)

= – (4/7 × 2/3 × 7/6) (a²b × b³c × c³a)

= – 4/9a³b³c³

Question no – (31)

Solution :

Given, (4/9abc³) × (- 27/5a³b²) × (-8b³c)

= (4/9 × 27/5³ × 8) (abc² × a³b×b³c)

= 96/5a⁴b⁶c⁴

Question no – (32)

Solution :

Given, (2xy) × (x²y/4) × (x²) ×(y²)

= 1/2x⁵y⁴

= 1/2(2)⁵ (1)² …[∵ x = 2, y = 1]

= 2⁴ × 1

= 16

Question no – (33)

Solution :

Given, (3/5x²y) × (-15/4xy²) × (7/9x²y²)

= – 7/4x⁵y⁵

= -7/4(2)⁵ (1)² [∵ x = 2, y = 1]

= -7/2 × 2 × 2 × 2 × 2 × 2 × 1

= -56

Algebraic Expressions and Identities Exercise 6.4 Solution :

Question no – (1)

Solution :

Given, 2a³(3a + 5b)

= (2a³ × 3a) + (2a³ × 5b)

= 6a⁴ + 10a³b

Question no – (2)

Solution :

-11a(3a + 2b)

= (-11a × 3a) + (-11a × 2b)

= -33a² – 22ab

Question no – (3)

Solution :

-5a(7a – 2b)

= (-5a × 7a) – (5a × 2b)

= 35a² + 10ab

Question no – (4)

Solution :

11y²(3y + 7)

= (- 11y² × 3y) + (-11y² × 7)

= -33y³ – 77y²

Question no – (5)

Solution :

6x/5 (x³ + y³)

= (6x/5 × x³) + (6x/5 × y³)

= 6x⁴/5 + 6xy³/5

Question no – (6)

Solution :

xy(x³ – y³)

= (xy × x³) – (6x/5 × y³)

= x⁴y – xy⁴

Question no – (7)

Solution :

0.1y (0.1x⁵ + 0.1y)

= (0.1x⁵ × 0.1y) + (0.1y × .01y)

= 0.01x⁵y + 0.01y²

Question no – (8)

Solution :

(-7/4ab²c – 6/25a²c²) (- 50a²b²c²)

= (-7/4ab²c × – 50a²b²c²) – (6/25a²c² × 50a²b²c²)

= 175/2a³b⁴c³ – 12a⁴b²c⁴

Question no – (9)

Solution :

-8/27 xyz (3/2xyz² – 9/4xy²z³)

= (- 8/27 xyz × 3/2xyz²) – (- 8/27 xyz× 9/4xy²z³)

= -4/9 x²y²z³ + 2/3 x²y³z⁴

Question no – (10)

Solution :

-4/27 xyz ( 9/2x²y z – 3/4 xyz²)

= (- 4/27xyz × 9/2x²yz) – (- 4/27xyz× 3/4xyz²)

= – 2/3x³y²z² + 1/9x²y²z³

Question no – (11)

Solution :

1.5x (10x²y – 100xy²)

= (1.5x × 10x²y) – (1.5x × 100xy²)

= 15x³y – 150x²y²

Question no – (12)

Solution :

4.1xy (1.x1x-y)

= (4.1xy × 1.1x) – (4.1xy ×y)

= 4.51x²y – 4.1xy²

Question no – (13)

Solution :

Given, 250.5xy(xz + y/10)

= (250.5xy × xz) – (250.5xy × x/10)

= 250.5x²yz + 25.05x²y

Question no – (14)

Solution :

= 7/5x²y (3/5xy² + 2/5x)

= (7/5x²y × 3/5xy²) + (7/5x²y × 2/5x)

= 21/25x³y³ + 14/25x³y

Question no – (16)

Solution :

24x² (1 – 2x)

= 24x² (1 – 2x)

= (24x² × 1) – (24x² × 2x)

= 24x² – 48x³

= 24 × 3² – 48 × 3³ [∵ x = 3]

= 24 × 9 – 48 × 27

= 216 – 1296

= – 1080

Question no – (17)

Solution :

= – 3y (xy + y²)

= (- 3y × xy) + ( -3y × y)

= – 3xy² – 3y³

= – 3× 4 × 5 – 3 × 5³

= – 300 – 375 [∵ x = 4, y = 5]

= – 675

Question no – (18)

Solution :

– 3/2 x²y³ (2x – y)

= (- 3/2 x²y³ × 2x) – (- 3/2 x²y³ × y)

= – 3x³ y³ + 3/2 v² y³ [∵ x = 1, y = 2]

= -3(1) ³ (2) ³ + 3/2 (1)² (2)⁴

= 0

Question no – (19)

Solution :

(i) (15y²(2 – 3x)

= (15y²× 2) – (15y²× 3x)

= 30y² -45xy²

= 30 × (0.25)² – 45 × (-1) × (0.25)² [ ∵ x = 1, y = 0.025]

= 30 × 0.0625+ 45 × 0.0625

= 1.875 + 2.8125

= 4.6875

(ii) – 3x (y²+ z²)

= – 3x y² – 3xz²

= -3 × (-1) × (0.25)² – 3 (-1) (0.05)²

= 3× 0.0625+ 3×0.0075 [∵ x = -1, y = 0.25, z = 0.05]

= 0.195

(iii) z² (x-y)

= z²+yz²

= (-1) (0.05)² – (0.25) (0.50)²

= -0.25-0 .0625 [∵ x = -1, y = 0.25, z = 0.0]

= 0.3125

(iv) xz(x² + y²)

= xz × x² + xz× y²

= x²z + xy²z

= (-1)³ × 0.25 + (-1) × (0.25)² × 0.50

= -1 × 0.25 – 0.03125 [∵ x = -1, y = 0.25, z = 0.50]

= 0.28125

Question no – (20)

Solution :

(i) 2x² (x³ – x) – 3x (x⁴ + 2x) – 2(x⁴ – 3x²)

= 2x⁵ – 2x³ – 3x⁵ – 6x² – 2x⁴ + 6x²

= -x⁵ – 2x⁴ – 2x³…(Simplified)

(ii) x³y(x² – 2x) + 2xy(x³ – x⁴)

= x⁵y – 2x⁴y + 2x⁴y – 2x⁵y

= – x⁵y…(Simplified)

(iii) 3a² + 2(a + 2) – 3a (2a + 1)

= 3a² + 2a + 4 – 6a² – 3a

= – 3a² – a + 4…(Simplified)

(vi) a(b – c) + b(c – a) + c(a – b)

= ab – ac + bc – ab + ac – bc

= 0…(Simplified)

(vii) 4ab (a-b) – 6a² (b – b²) – 3b²(2a² – a) + 2ab(b – a)

= 4a²b – 4ab² – 6a²b + 6a²b² – 6a²b² + 3ab² + 2ab² – 2a²b

= (4 – 6 – 2) a²b – (4 – 2 + 3)

= – 4a²b + ab²…(Simplified)

(ix) 2a² + 3a (1 – 2a³) + a(a + 1)

= 2a² + 3a – 6a⁴ + a² + a

= 3a² + 4a – 6a⁴…(Simplified)

(x) a² (2a – 1) + 3a + a³ – 8

= 2a³ – a² + 3a + a³ – 8

= 3a³ + 3a – a² – 8…(Simplified)

(xi) 3/2x² (x² – 1) + 1/4 x² (x² + x) – 3/4x (x³ – 1)

= 3/2x⁴ – 3/2x² + 1/4x⁴ + 1/4x³ – 3/4x⁴ + 3/4x

= (3/2 + 1/4 – 3/4) x⁴ – 3/2x² + 1/4x³ + 3/4x

= x⁴ + 1/4x³ – 3/2x² + 3/4x…(Simplified)

(xii) a²b (a – b²) + ab² (4ab – 2a²) – a³b(1 – 2b)

= a³b – a²b³ + 4a²b³ – 2 a³b² – a³b + 2a³b²

= 3 a²b³…(Simplified)

(xiii) a²b (a³ – a + 1) – ab (a⁴ – 2a² + 2a) – b(a³ – a² – 1)

= a⁵b – a³b + a²b – a⁵ b + 2a³b – 2a²b -a³b + a² b + b

= a³b + a³b + a²b – a²b + b

= b…(Simplified)

Algebraic Expressions and Identities Exercise 6.5 Solution :

Question no – (1)

Solution :

Given, (5x + 3) by (7x + 2)

= 5x(7x + 2) + 3 (7x + 2)

= 35x² + 10x + 21x + 6

= 35x² + 31x + 6

Question no – (2)

Solution :

Given, (2x + 8) by (x – 3)

= 2x (x – 3) + 8(x – 3)

= 2x² – 6x + 8x – 24

= 2x² + 2x – 24

Question no – (3)

Solution :

Given, (7x + y) by (x + 5y)

= 7x (x + 5y ) + y(x+5y)

= 7x² + 35y + xy + 5y²

= 7x² + 36xy + 5y²

Question no – (4)

Solution :

Given, (a – 1) by (0.1a² + 3)

= a (0.1a²) – 1 (0.1² + 3)

= 0.1a³ + 3a – 0.1a² – 3

Question no – (5)

Solution :

Given, (3x² + y²) by (2x² + 3y²)

= 3x² (2x² + 3y²) + y²(2x² + 3y²)

= 6x⁴ + 9x²y² + 2x²y² + 3y⁴

= 6x⁴ + 11x²y² + 3y⁴

Question no – (6)

Solution :

Given, (3/5x + 1/2y) by (5/6x + 4y)

= 3x/5 (5/6x +4y) + 1/2y (5/6x + 4y)

= 1/2x² + 12/5xy + 5/12 + 2y²

= 1/2x² + 169/60xy + 2y²

Question no – (7)

Solution :

Given, (x⁶ – y⁶)by (x² + y²)

= x⁶ (x² + y²) – y⁶(x² + y²)

= x⁸ + xy² – y⁶ x² – y⁸

Question no – (8)

Solution :

Given, (x² +y²) by (3a + 2b)

= x²(3a+2b) + y²(3a+2b)

= 3x²a + 2x²b + 3ay² + 2by²

Question no – (9)

Solution :

[-3d + (7f)] by (5d + f)

= -3d (5d + f) + -7f (5d + f)

=-15d² – 3df – 35df + 7f²

= -15d² – 38df + 7f²

Question no – (10)

Solution :

(0.8a – 0.5b) by (1.5a-3b)

= 0.8a (15a-3b) – 0.5b (0.5a -43b)

= 0.2a² – 2.4ab – 0.75ab + 1.5b²

= 1.2a² – 3.15ab + 1.5b²

Question no – (11)

Solution :

As per the question, (2x²y² – 5xy²) by (x² – y²)

= 2x²y²(x² – y²) – 5xy² (x² + y²)

= 2x⁴y² – 2x⁴y² – 5x⁴y² + 5xy⁴

Question no – (12)

Solution :

Given,  (x/7 + x²/2) by (2/5 + 9x/4)

= x/7 (2/5 + 9x/4) + x²/2 (2/5 + 9x/4)

= 2x/35 + 9/28x² + 1/5x² + 9/8x³

= 2/35 x + (45 + 28/14)x² + 9/8x³

= 9/8x³ + 73/140x² + 2/35x

Question no – (13)

Solution :

Given in the question, (-a/7+ a²/9) by (b/2 – b²/3)

= -a/47 (b/2 – b²/3) + a²/9 (b/2 – b²/3)

= -ab/14 + ab²/21 + a²b/18 – a²b²/27

Question no – (14)

Solution :

As per the question, (3x²y – 5xy²) by (1/5x² + 1/3y²)

= 3x²y (1/5x² + 1/3y²) – 5xy²(1/5x² + 1/3y²)

= 3/5x⁴y + x²y³ – x³y² – 5/3xy⁴

Question no – (15)

Solution :

Given, (2x² -1) by (4x³ + 5x²)

= 2x² (4x³ + 5x²) -1(4x³ +5x²)

= 8x⁵ + 10x⁴ – 4x³ – 5x²

Question no – (16)

Solution :

Given,  (2xy+3y²) (3y²-2)

= 2xy (3y²-2) + 3y² (3y²-2)

= 6xy³ – 4xy + 9y⁴ – 6y²

Question no – (17)

Solution :

Given, (3x – 5y) (x + y)

= 3x (x + y) – 5y (x + y)

= 3x² + 3x – 5xy – 5y²

= 3x² – 2xy – 5y²

Therefore, the product will be 3x² – 2xy – 5y²

Question no – (18)

Solution :

Given, (x²y – 1) (3 – 2x²y)

= x²y(3 – 2x²y) – 1 (3 – 2x²y)

= 3x²y – 2x⁴y² – 3 + 2x²y

= 5x²y – 2x⁴y² – 3

Hence, the product will be 5x²y – 2x⁴y² – 3

Question no – (19)

Solution :

Given, (1/3x – y²/5) (1/3x + y²/5)

= 1/9x²(1/3x + y²/5) – y²/5(1/3x + y²/5)

= 1/9x² – xy² /15 – xy²/15 – y⁴/25

= 1/9x² – y⁴/25

Hence, the product will be 1/9x² – y⁴/25

Simplify.

Question no – (20)

Solution :

Given, x²(x + 2y) (x – 3y)

= x²[x (x – 3y) + 2y (x – 3y)]

= x²[x² – 3xy + 2y – 6y²]

= x²[x² – xy – 6y²]

= x⁴ – x³y – 6x²y²…(Simplified)

Question no – (21)

Solution :

Given, (x² – 2y²) (x + 4y)x²y²

= [x² (x+4y) -2y² (x+4y)] x²y²

= [x³ + 4x²y – 2xy² – 8y] x²y²

= x⁵y² + 4x⁴y³ – 2x³y⁴ – 8x²y⁵…(Simplified)

Question no – (22)

Solution :

Given, a²b² (a + 2b) (3a + b)

= a²b² [a(3a + b) + 2b (3a + b)]

= a²b² [3a² + ab + 6ab + 2b²]

= a²b² [3a² + 7ab + 2b²]

= 3a⁴b² + 7a³b³ + 2a²b⁴…(Simplified)

Question no – (23)

Solution :

Given, x²(x – y) y²(x + 2y)

= (x³ – x²y) (xy² + 2y³)

= x³ (xy² + 2y³) – x²y (xy² + 2y³)

= x⁴y² + 2x³y³ – x³y³ – 2x²y⁴

= x⁴y² + x³y³ – 2x²y⁴…(Simplified)

Question no – (24)

Solution :

Given, (x³ – 2x² + 5x – 7) (2x – 3)

= 2x(x³ – 2x² + 5x – 7) – 3(x³ – 2x² + 5x – 7)

= 2x – 4x³ + 10xy – 11x – 3x³ + 6x² – 15x + 21

= 2x – 7x³ + 16x² – 29x + 21…(Simplified)

Question no – (25)

Solution :

Given, (5x + 3) (x – 1) (3x – 2)

= (5x + 3) [x (3x – 2) -1 (3x – 2)]

= (5x + 3) (3x² – 2x – 3x + 2)

= (5x + 3) (3x³ – 5x + 2)

= 5x (3x² – 5x + 2) + 3(3x² – 5x + 2)

= 15x³ – 25x² + 10x + 15 + x² – 15x + 6

= 15x³ – 10x² – 5x + 6…(Simplified)

Question no – (27)

Solution :

Given, (2x² + 3x – 5) (3x² – 5x + 4)

= 2x² (3x² – 5x +4) + 3x (3x² – 5x+4) – 5 (3x² – 5x + 4)

=6x⁴ – 10x³ + 8x + 9x³ -15 + 12x – 15x² + 25x – 20

= 6x⁴ – x³ – 22x² + 9x² + 37x – 20…(Simplified)

Question no – (28)

Solution :

Given, (3x – 2) (2x – 3) + (5x – 3) (x + 1)

= 6x² – 9x – 4x + 6 + 5x² + 5x – 3x – 3

= 11x² – 11x + 3…(Simplified)

Question no – (30)

Solution :

Given, (3x + 2y) (4x + 3y) – (2x – y) (7x – 3y)

= (12x² + 9xy + 8xy + 6y²) – (14x² – 6xy – 7xy + 3y²)

= 12x² + 17xy + 6y² – (14x² – 13xy + 3y²)

= 12x² + 17xy + 6y² – 14x² +13y – 3y²

= – 2x² + 30 xy + 3y²…(Simplified)

Question no – (32)

Solution :

Given, (x³ -2x² + 3x – 4) (x-1) – (2x – 3) (x² – x + 1)

= (x⁴ – 2x³ + 3x² – 4x – x³ – 2x² – 3x + 4) – (2x³ – 2x² + 2x – 3x² + 3x – 3)

= (x⁴ – 3x³ + 5x² – 7x + 4 ) – ( 2x³ – 5x² + 5x – 3)

= x⁴ – 3x³ + 5x² – 7x + 4 – 2x³ + 5x² – 5x + 3

= x⁴ – 5x³ + 10x² – 12x + 7…(Simplified)

Algebraic Expressions and Identities Exercise 6.6 Solution :

Question no – (1)

Solution :

(i) (x + 2)²

= x² + 2.2.x + 2²

= x² + 4x + 4

(ii) (8a + 3b)²

= (8a)² + 2.8a – 3b + (3b)²

= 64a² + 48ab + 9b²

(iv) (9a + 1/6)²

= (9a)² +2.9a.1/6 + (1/6)²

= 81a² + 3a + 1/36

(v) (x + x²/2)²

= x² + 2. x. x²/2 + (x²/2)

= x² + x³ + x⁴/4

(vi) (x/4 – y/3)

= (x/4)² + 2 .x/4. y/3 + (b/3) ²

= x²/16 + xy/6 + y²/9

(vii) (3x – 1/3x)²

= (3x)² + 2. 3x. 1/3x + (1/3x)²

= 9x² + 2 + 1/9x²

(viii) (x/y – x/y)²

= (x/y)² + 2. x/y. y/x + (y/x) ²

= x²/y² + 2 + y²/x²

(ix) (3a/2 – 5b/4)²

= (3a/2) ² +2. 3a/2 . 5b/2 + (5b)²

= 9a²/4 + 15ab/2 + 25b²

(x) (a²b – bc²)²

= (a² b²)² + 2. a²b + bc² + (bc²)²

= a⁴ b² + 2a²b² + b²c⁴

(xi) (2a/3b + 2b/3a)²

= (2a/3b)² + 2. 2a/3b .2b/3a + (2b/3a) ²

= 4a²/9b² + 8/9 + 4b²/9a²

Question no – (2)

Solution :

(i) (2x + y) (2x + y)

= (2xy)²

= (2x)² + 2. 2x y + y²

= 4x² + 4xy + y²

Hence, the product will be 4x² + 4xy + y²

(ii) (a + 2b) (a – 2b)

= a² – (2b)²

= a² – 4b²

Thus, the product will be a² – 4b²

(iii) (a² + bc ) (a² – bc)

= (a²)² – (bc)²

= a⁴ – b²c²

Therefore, the product will be a⁴ – b²c²

(iv) (4x/5 – 3y/4) (4x/5 + 3y/4)

= (4x/5)² – (3y/4)²

= 16x²/25 – 9y²/16

Therefore, the product will be 16x²/25 – 9y²/16

(v) (2x + 3/y) (2x – 3/y)

= (2x) ² – (3/y) ²

= 4x² – 9/y²

Thus, the product will be 4x² – 9/y²

(vi) (2a³ + b³) (2a³ – b³)

= (2a³)² – (b³)²

= 4a⁶ – b⁶

Hence, the product will be 4a⁶ – b⁶

(vii) (x⁴ + 2/x²) (x⁴ – 2/x²)

= (x⁴)² – (2/x²)

= x⁸ – 4/x⁴

Thus, the product will be x⁸ – 4/x⁴

(viii) (x³+ 1/x³) (x³ – 1/x³)

= (x³)² – (1/x³)²

= x⁶ – 1/x⁶

Therefore, the product will be x⁶ – 1/x⁶

Question no – (3)

Solution :

(i) (100)²

= (100 + 10²

= (100) ² + 2.100. 1 + 1²

= 10000 + 200 + 1

= 10201

(ii) (99)²

= (100 – 1)²

= (100)² – 2.1000.1 + 1²

= 10000 – 200 + 1

= 9801

(iii) (1001)²

= (100 + 1)

= (1000)² + 2.100 + 1 + 1²

= 1000000 + 2000 + 1

= 10020001

(iv) (999)²

= (100-1)²

= (1000)² – 2.1000 + 1 + 1²

= 1000000 – 2000 + 1

= 998001

(v) (700 + 3)²

= (700 + 3)²

= (700)² + 2.700 + 3 + 3²

= 490000 + 4200 + 9

= 532009

Question no – (4)

Solution :

(i) (82)² – (18) ²

= (82 + 18) (82 – 18)

= 100 × 64

= 6400…(Simplified)

(ii) (467)² – (33)²

= (467 + 33) (467 – 33)

= 500 × 434

= 217000…(Simplified)

(iii) (79) ² – (69) ²

= (79 + 69) (79 – 69)

= 148 × 10

= 1480…(Simplified)

(iv) 197 × 203

= (200 – 3) (200 + 3)

= (200)² – 3²

= 40000 – 9

= 39991…(Simplified)

(v) 113 × 87

= (100 + 13) (100 – 13)

= (100)² – (13)²

= 1000 – 169

= 9831…(Simplified)

(vi) 95 × 105

= (100 – 5) (100 + 5)

= (100)² – 5² – 1000 – 25

= 9975…(Simplified)

(viii) 9.8 × 10.2

= (10.0 – 0.2) (10.0 + 0.2)

= (10.0)² – (0.2)²

= 100.00 – 0.04

= 99.96…(Simplified)

Question no – (5)

Solution :

(i) 58² – 42²/16

= (58 + 42) (58 – 42)/16

= 100 × 16/16

= 100…(Simplified)

(ii) 178 × 178 – 22 × 22

= (178)² – (22)²

= (178 × 22) (178 – 22)

= 200 × 156

= 31200…(Simplified)

(iii) 198 × 198 – 102 × 102/96

= (198)² – (102)²/96

= (198 + 102) (198 – 102)/96

= 300 × 96/96

= 300…(Simplified)

(iv) 1.73 × 1.73 – 0.27 × 0.27

= (1.73)² – (0.27)²

= (1.73 + 0.27) (1.73 – 0.27)

= 2.00 × 1.46

= 2.92    …(Simplified)

(v) 8.63 × 8.63- 1.37 × 1.37/0.726

= (8.63)² – (1.37)² /0.726

= (8.63 + 1.37) (8.63 – 1.37)/0.726

= 10.00 × 7.26/0.26

= 100.0 × 7.26/10 × 0.726

= 100 × 7.26/7.26

= 100    …(Simplified)

Question no – (6)

Solution :

(i) 4x = (52) ² – (48) ²

or 4x = (52 + 48) (52 – 48)

or 4x = 100 × 4

or x = 100 × 4/4

x = 100

Therefore, the value of x will be 100

(ii) 14x = (47)² – (33)²

or, 14x = (47 + 33) (47 – 33) = 80 × 14

or x = – 80 × 14/14 = 80

Thus, the value of x will be 80

(iii) 5x = (50)² – (40)²

Or, 5x = (50 + 40) (50 – 40) = 90 × 10

Or, x = 90 × 10/5 = 180

Hence, the value of x will be 180

Question no – (7)

Solution :

Given, x + 1/x = 20

or (x + 1/x)² = (20)²

or x² + 2x 1/x + 1/x² = 400

or x² + 1/x² + 2 = 400

or x² + 1/x² = 400 – 2 = 398

or x² + 1/x² = 398

Hence, the value will be 398

Question no – (8)

Solution :

Give, x – 1/x = 3

or, (x- 1/x)² = 32

or, x²-2 x. 1/x + (1/x)² = 9

or, x² + 1/x² – 2 = 9

or, x² + 1/x² = 9 + 2 = 11

(x² + 1/x²)² = (11)² …… (Squaring both the term)

= x² + 2 . x . 1/x² + 1/x⁴ = 121

= x⁴ + 1/x⁴ = 121 – 2

= x⁴ + 1/x⁴ = 119

Therefore, the values are 11 and 119

Question no – (9)

Solution :

As per the question, x² + 1/x² = 18

or, x² + 1/x² + 2 – 2= 18

or, x² + 2. x. 1/x + 1/x² – 2 = 18

or, (x + 1/x)² = 18 + 2 = 20

or, (x + 1/x) = √20

Again,

x² + 1/x² = 10

or, x² + 1/x² + 2 – 2= 18

or, x² – 2. x 1/x + 1/x² + 2 = 18

or, (x – 1/x)² = 18 – 2 = 16

or, x – 1/x = √16 = 4

x + 1/x = √20

x – 1/x = 4

Therefore, the values are √20 and 4.

Question no – (10)

Solution :

Given,  x + y = 4; xy = 2

or, (x + y)² = 4

or x² + 2x + 2xy +y² = 16

or, x² + y² + 2.2 = 16

or, x² + y² +4 = 16

or, x² + y²

= 16 – 4

= 12

Therefore, the value of x² + y² will be 12.

Question no – (11)

Solution :

Given, x – y = 7

Squaring both side,

= (x – y)² = 7²

= x² + y² – 2xy = 49

= x² + y² = 49 + 2xy

Now,

= x² + y² = 49 + 2 × 9

= (49 + 18)

= 67

Hence, the value of (x² + y²) will be 67

Question no – (12)

Solution :

As per the question, 3x + 5y = 11; xy = 2

or, (3x + 5y)² = 11²

or, (3x) ² + 2. 3x. 5y + (5y)² = 121

or, 9x² + 30xy + 25 y² + 30.2 = 121

or, 9x² + 25y² = 121 – 60 = 61

or, 9x² + 25y² = 61

Therefore, the value will be 61

Question no – (13)

Solution :

(i) 16x² + 24x + 9

= (4x)² + 2. 4x .3 + 3²

= (4x + 3)² = (4. 7/4 + 3) ….[∵ x = 7/4]

= (7 + 3)²

= (10)² = 100

Thus, the value will be 100

(ii) 64x² + 81y² + 144xy

= (8x) ² + (9y) + 2. 8x. 9y

= (8x + 9y) (8.11 + 9.4/3) ….[∵ x = 11, y = 4/3]

= 988 + 12)² = (100)²

= 10000

Therefore, the value will be 10000

(iii) x = 2/3; y = 3/2

81x² + 16y² – 72xy

= 81x² – 72xy + 16y²

= (9x)² – 2. 9x. 4y + (4y)²

= (9x – 4y)² = (9 .2/3 – 4 . 3/4)²

= (6 – 3)² = 3² = 9

Therefore, the value will be 9

Question no – (14)

Solution :

Given value, x + 1/x = 9

x + 1/x = 9

or (x + 1/x) = 9²

or, x² + 2.x 1/x + (1/x)²= 81

or x² +1/x² + 2 = 81

or, x²1/x² = 81 – 2 = 79

or, (x² + 1/x²)² = (79)²

or (x²) + 2. x²1/x² + (x²)² = 6241

or, x⁴ + 1/x ⁴ + 2 = 6241

or, x⁴ + 1/x⁴

= 6241 – 2

= 6239

Therefore, the value will be 6239

Question no – (16)

Solution :

As per the question,

2x + 3y = 14; 2x – 3y = 2; xy = ?

(2x + 3y) ² – (2x-3y)² = 24xy

or (14)² – 92)² = 24xy

or, 196 – 4 = 24xy

or, xy = 192/24 = 8

Therefore, the value of xy will be 8

Question no – (17)

Solution :

(i) x² + y² = 29; xy = 2

x² + y² = 29

or, (x – y)² – 2xy = 29

or, (x + y)² 2. 2 = 29

or, (x + y) = 29 + 4 = 33

or x + y = ±√33

(ii) x² + y² = 29; xy = 2

x² + y² = 29

or, (x – y) ² – 2xy = 29

or, (x – y)² + 2.2 = 29

or, (x – y)² = 29 – 4 = 33

or, x – y = √ = ±5

Therefore, the value of x – y will be ±5

(iii) x² + y² = 29; xy = 2

x² + y² = 29

or (x² + y²)² = (29)²

or (x²)² + 2. x²y² + y⁴ = 841

or, x⁴ + y⁴ + 2.4 = 841

or, x⁴ + y⁴ = 841 – 8 = 833

Hence, the value of x⁴ + y⁴ = 833

Question no – (18)

Solution :

(i) 4x² – 12x + 7

= (2x)²- 12 + 12x + 7 + 2

= (2x)² – 2.2x .3 + (3)²

= (2x) – 2. 2x 3 + (3)²

= (2x + 3)²

(ii) 4x² – 20x + 20

= (2x)² – 20x + 20 + 5

= (2x)² – 20x + 25

= (2x)² 2 . 2x .5 + 5²

= (2x + 5)²

Question no – (19)

Solution :

(i) (x – y) (x + y) (x² + y²) (x⁴ + y⁴)

= (x² – y²) (x² + y²) (x⁴+ y⁴)

= {(x²)² – (y²)²} (x⁴ + y⁴)

= (x⁴- y⁴) (x⁴ + y⁴)

= (x⁴)² – (x⁴)²

= x⁸ – y⁸…(Simplified)

(ii) (2x – 1) (2x + 1) (4x² + 1) (16x⁴ + 1)

= [(2x)² – 1²] (4x² + 1) (16x + 1)

= (4x² – 1) (4x² + 1) (16x⁴ + 1)

= [(4x²)² – (1²)² ] [16x⁴ + 1]

= (16x⁴ – 1) (16x⁴ + 1) = (16x⁴) – 1²

= 256x⁸ – 1…(Simplified)

(iii) (7m – 8n)² + (7m + 8n)²

= (7m)² – 2.7m 8.n + (8n)² + (7m)² + 2.7m .8n + (8n)²

= 49m² + 49m² + 64n² + 64n²

= 98m² + 128n²…(Simplified)

(iv) (2.5p – 1.5q)² – (1.5 – 2.5q)²

= [(2.5p)² – 2 × 2. 5p × 1. 5q +(1.5q)²] – [(1.5p) – 2. 1. 5p. 2. 5q + (2.5q)²]

= 6.25p² – 7.5 pq + 2.25q² + 2. 25p² + 7.5pq – 6.25q²

= 4p² – 4q² = 4(p² – q²)

= 4(p² – q²)…(Simplified)

(v) Given, (m² – n²m) ² + 2m³n²

= (m²)² – 2.m². n².m + (n²m)² + 2m³n²

= m⁴ – 2m³n² + n⁴m² + 2m³n²

= m⁴ + n⁴m²…(Simplified)

Question no – (20)

Solution :

(i) L.H.S, (3x + 7)² – 84x

= (3x + 7) – 84x

= (3x)² + 2. 3x .7 + 7² – 84x

= (3x)² + 42x + (7)² – 84x

= (3x)² – 42x + (7)²

= (3x – 7)²

L.H.S = R.H.S    …[proved]

(ii) L.H.S (9a-5b)² + 180ab

= (9a)² – 2. 9a. 5b + (5b)² + 180ab

= (9a)² – 90ab + (5b) ² + 180 ab

= (9a)² + 90ab + (5b)² = (9a + 5b)²

L.H.S = R.H.S   …[proved]

(iii) L.H.S (4m/3 – 3n/4)² + 2mn

= (4m/3)² – 2 . 4m/3 . 3n/4 + (3n/4)² + 2mn

= (4m/3)² – 2mn + 2mn + (3n/4)²

= (4m/3)² + (3n/4)² = 16m²/9 + 9n²/16

L.H.S = R.H.S   …[proved]

(iv) L.H.S (4pq + 3q)² – (4pq – 3p)

= (4pq)² + 2.4 pq. 3a + (3q)² – [(4pq)² – 2. 4 pq. 3q + (3q)²

= (4pq)² + 24pq + (3q)² – (4pq)² + 24pq – (3q)² = 48 pq²

L.H.S = R.H.S …..[proved]

Algebraic Expressions and Identities Exercise 6.7 Solution :

Question no – (1)

Solution :

(i) (x + 4) (x + 7)

= x² + (4 + 7) x + 4.7

= x² + 11x + 28

Thus, the product will be x² + 11x + 28

(ii) (x – 11) (x + 4)

= [x + (-11)] [x + 4]

= x² + (-11 + 4) x + (- 11). 4

= x² + (-7) x + (- 44)

= x² – 7x – 44

Hence, the product will be x² – 7x – 44

(iii) Given, (x + 7) (x – 5)

= [(x + 7)] [x + (-5)]

= x²[7 + (-5)] x + 7.(-5)

= x² + (2) – 35

= x² + 2x – 35

So, the product will be x² + 2x – 35

(iv) (x – 3) (x – 2)

= x² – (3 + 2) x + 3.2

= x² – 5x + 6

Thus, the product will be x² – 5x + 6

(v) (y² – 4) (y² – 3)

= (y²)² – (4 + 3) y² + 4.3

= 4² – 7y² + 12

Therefore, the product will be 4² – 7y² + 12

(vi) (x + 4/3) (x + 3/4)

= x² + (4/3 + 3/4) x + 4/3 . 3/4

= x² + (16 + 9/12) x + 1

= x² + 25/12x + 1

Hence, the product will be x² + 25/12x + 1

(vii) Let, 3x = y

(y + 5) (y + 11) = y² + (5 + 11) y + 5. 11

= y² + 16y + 55

= (3x)² + 16. 3x + 55 …[Putting value of y = 3x]

= 9x² + 48x + 55

Thus, the product will be 9x² + 48x + 55

(viii) Let, 2x² = y

(y – 3) (y + 5) = (y + 5) (y – 3)

= [y + 5] [y + (-3)]

= y² + [5 + (-3)] y 5 × (-3)

= y² + 2y – 15

= (2x²)² + 2. 2x² – 15 ….[∵ y = 2x²]

= 4x⁴ + 4x⁴ – 15

So, the product will be 4 x⁴ + 4x⁴ – 15

(ix) Let , z²= x

(x + 2) (x – 3) = [x + 2] [x + (- 3)]

= x² + [2 + (-3)] x + 2. (-3)

= x² – x – 6 = (z²) – z² – 6

= x² – z² – 6

Thus, the product will be x² – z² – 6

(x) (3x – 4y) (2x – 4y)

= 3x. 2px – 3x. 4y – 4y – 2x + 4y. 4y

= 6x² – 12xy – 8xy + 16y²

= 6x² – 20xy + 16y²

Hence, the product will be 6x² – 20xy + 16y²

(xi) (3x² – 4xy)(3x² – 3xy)

= (3x²) – (4xy + 3xy) 3x² + (4xy) . (3xy)

= 9x⁴ – 7xy. 3x² + 12x²y²

= 9x⁴ – 21x³y + 12x²y²

Therefore, the product will be 9x⁴ – 21x³y + 12x²y²

(xii) (x + 1/5) (x + 5)

= x² + (1/5 + 5) x + 1/8. 5

= x² + (1 + 25) x + 1

= x² + 26/5 x + 1

Hence, the product will be x² + 26/5 x + 1

(xiii) (z + 3/4) (z + 4/3)

= z² + (3/4 + 4/3) z + 3/4 . 4/3

= z² + (9 + 16/12) z + 1

= z² + 25/12 z + 1

Thus, the product will be z² + 25/12 z + 1

(xiv) (x² + 4) (x² + 9)

= (x²)² + (4 + 9) x² +4 × 9

= x⁴ + 13x² + 36

So, the product will be x⁴ + 13x² + 36

(xv) (y² + 12) (y² + 6)

= (y²)² + (12 + 6) y² + 12 × 6

= y⁴ + 18y² + 72

Thus, the product will be y⁴ + 18y² + 72

(xvi) (y² + 5/7) (y² – 14/5)

= (y² + 5/7) [ y² + (-14/5)]

= (y²)² – [5/7 + (-14/5)] y² + 5/7 × (-14/8)²

= y⁴ – [25 – 98/35] y² – 2

= y⁴ – 73/35 y² – 2

Therefore, the product will be y⁴ – 73/35 y² – 2

(xvii) (p² + 16) (p² – 1/4 )

= [p² + 16] [p² + (-1/4)]

= (p²)² + [16 + (-1/4)] p² + 16 × (-1/4)

=p⁴ + [ 64 – 1/4] p² – 4

= p⁴ + 63/4 p² – 4

Therefore, the product will be p⁴ + 63/4 p² – 4

Question no – (2)

Solution :

(i) 102 × 106

= (100 + 2) (100 + 6)

= (100)² + (2 + 6) 100 + 2.6

= 10000 + 800 + 12

= 10000 + 800 + 12

(ii) 109 × 107

= (100 + 9) (100 + 7)

= (100)² + (9 + 7) 100 + 9.7

= 1000 + 16 × 100 + 63

= 10000 + 1600 + 63

(iii) (30 + 5) (30 + 7)

= (30)² + (5 + 7) 30 + 5.7

= 900 + 12 × 30 + 35

= 900 + 360 + 35

(iv) 53 × 55

= (50 + 3) (50 + 5)

= (50)² + (3 + 5) 50 + (3 × 5)

= 2500 + 8 × 50 + 15

= 2500 + 400 + 15

(v) 103 × 96

= (100 + 3) (100-4)

= [100 + 3] [100 + (4)]

= (100)² – ( 3 + (- 4)] 100 + 3. (-4)

= 1000 + 100 – 12

(vi) 34 × 36

= (40 – 6) (40 – 4)

= (40)² – (6 + 4) 40 + 6.4

= 1600 – (10 × 40) + 24

= 1600 – 400 + 24

(vii) 994 × 1006

= (100 – 06) (100 + 6)

= (1000)² – 6²

= 100000000 – 36