Class 8 ICSE Maths Solutions Chapter 12 Identities (Selina Concise)
Welcome to NCTB Solutions. Here with this post we are going to help 8th class students for the Solutions of Selina Class 8 ICSE Math Book, Chapter 12, Identities. Here students can easily find step by step solutions of all the problems for Identities, Exercise 12A, 12B, 12C and 12D Also here our mathematics teacher’s are solved all the problems with easily understandable methods with proper guidance so that all the students can understand easily. Here in this post students will get chapter 12 solutions. Here in this post all the solutions are based on ICSE latest Syllabus.
Identities Exercise 12(A) Solution :
Question no – (1)
Solution :
(i) (x + 8) (x + 3)
= x² + 3x + 8x + 24
= x² + 11x + 24
(ii) (y + 5) (y – 3)
= y² – 3y + 5y – 15
= y² + 2y – 15
(iii) (a – 8) (a + 2)
= a² + 2a – 8a – 16
= a² – 6a + 16
(iv) (b – 3) (b – 5)
= b² – 5b – 3b + 15
= b² – 8b + 15
(v) (3x – 2y) (2x + y)
= 6x² + 3xy – 4xy – 2y²
= 6x² – xy – 2y²
(vi) (5a + 16) (3a – 7)
= 15a² – 35a + 48a – 112
= 15a² + 13a – 112
(vii) (8 – b) (3 + b)
= 24 + 8b – 3b – b²
= 24 + 5b + b²
Question no – (2)
Solution :
(i) (x + 1) (x – 1)
= x² – 1²
= x² – 1
(ii) (2 + a) (2 – a)
= (2)² – (a)²
= 4 – a²
(iii) (3b – 1) (3b + 1)
= (3b)² – (1)²
= 9b² – 1
(iv) (4 + 5x) (4 – 5x)
= (4)² – (5x)²
= 16 – 25x²
(vi) (xy + 4) (xy – 4)
= (xy)² – (4)²
= x²y² – 16
(vii) (ab + x²) (ab – x²)
= (ab)² – (x)²
= a²b² – x⁴
(viii) (3x² + 5y²) (3x² – 5y²)
= (3x²)² – (5y²)²
= 9x⁴ – 25y⁴
(ix) (z – 2/3) (z + 2/3)
= (z)² – (2/3)²
= z² – 4/9
(x) (3/5a + 1/2) (3/5a – 1/2)
= (3/5a)² – (1/2)²
= 9/15a² – 1/4
(xi) (0.5 – 2a) (0.5 + 2a)
= (0.5)² – (2a)²
= 0.25 – 4a
(xii) (a/2 – b/3) (a/2 + b/3)
= (a/2)² – (b/3)²
= a²/4 – b²/9
Question no – (3)
Solution :
(i) Given, (a + 1) (a – 1) (a² + 1)
= (a² – 1) (a² + 1)
= (a²)² – (1)²
= a⁴ – 1
(ii) (a + b) (a – b) (a² + b²)
∴ (a² – b²) (a² + b²)
= (a²)² – (b²)²
= a⁴ – b⁴
(iii) (2a – b) (2a + b) (4a² + b²)
∴ (2a)² – (b)² (4a² + b²)
= (4a² – b²) (4a² + b²)
= (4a²)² – (b²)²
= 16a⁴ – b⁴
(iv) (3 – 2x) (3 + 2x) (9 + 4x²)
∴ (3)² – (2x)² (9 + 4x²)
= (9 – 4x²) (9 + 4x²)
= (9)² – (4x²)²
= 81 – 16x⁴
(v) (3x – 4y) (3x + 4y) (9x² + 16y²)
∴ (3x)² – (4y)² (9x² + 16y²)
= (9x² – 16y²) (9x² + 16y²)
= (9x²)² – (16y²)²
= 81x⁴ – 256y⁴
Question no – (4)
Solution :
As per the question, (a + b) (a – b) = a² – b²
(i) Given, 21 × 19
= (20 + 1) (20 – 1)
= (20)² – (1)²
= 400 – 1
= 399
(ii) Given, 33 × 27
= (30 + 3) (30 – 3)
= (30)² – (3)²
= (900 – 9)
= 891
(iii) In the question, 103 × 97
= (100 + 3) (100 – 3)
= (100)² – (3)²
= 1000 – 9
= 9991
(iv) 9.8 × 10.2
= (10 – 0.2) (10 + 0.2)
= 100 – 0.04
= 99.96
(v) 7.7 × 8.3
= (8 – 0.3) (8 + 0.3)
= (8)² – (0.3)²
= 64 – 0.09
= 63.91
(vi) 4.6 × 5.4
= (5 – 0.4) (5 + 0.4)
= (5)² – (0.4)²
= 25 – 0.16
= 24.84
Question no – (5)
Solution :
(i) Given, (6 – xy) (6 + xy)
∴ (6)² – (xy)²
= 36 – x²y²
(ii) (7x + 2/3y) (7x – 2/3y)
∴ (7x)² – (2/3y)²
= 49x² – 4/9y²
(iii) (a/2b + 2b/a) (a/2b – 2b/a)
∴ (a/2b)² – (2b/a)²
= a²/4b² – 4b²/a²
(iv) (3x – 1/2y) (3x + 1/2y)
∴ (3x) ² – (1/2y) ²
= 9x² – 1/4y²
(v) (2a + 3) (2a – 3) (4a² + 9)
∴ (2a) – (3) (4a² + 9)
= (4a² – 9) (4a² + 9)
= (4a²)² – (9)²
= 16a⁴ – 81
(vi) (a + bc) (a – bc) (a + b²c²)
∴ (a – b²c²) (a + b²c²)
= (a²)² – (b²c²)²
= a⁴ – b⁴c⁴
(vii) (5x + 8y) (3x + 5y)
= 15x² + 25y + 24xy + 40y²
= 15x² + 49xy + 40y²
(viii) (7x + 15y) (5x – 4y)
= 35x² – 28xy + 75xy – 60y²
= 35x² + 47xy – 60y²
(ix) (2a – 3b) (3a + 4b)
= 6a² – 8ab – 9ab – 12b²
= 6a² – ab – 12b²
(x) (9a – 7b) (3a – b)
= 27a² – 9ab – 21ab + 7b²
= 27a² – 30ab + 7b²
Identities Exercise 12(B) Solution :
Question no – (1)
Solution :
(i) Given, (2a + b)²
= (2a)² + 2.2a. b + b²
= 4a² + 4ab + b²
(ii) (a – 2b)²
= a² – 2.a. 2b + (2b) ²
= a² – 4ab + 4b²
(iii) (a + 1/2a)²
= (a)² + 2.a.1/2a + (1/2a)²
= a² + 1 + 1/4a²
(iv) (2a – 1/a)²
= (2a)² – 2.2a.1/a + (1/a) ²
= 4a² – 4 + 1/a²
(v) (a + b – c)²
= a² + b + (- c)² + 2ab – 2bc – 2ca
= a² + b² + c² + 2ab – 2bc – 2ca
(vi) (a – b – c)²
= a² + b² + c² – 2ab – 2bc + 2ca
(vii) (3x + 1/3x)²
= (3x)² + 2.3x.1/3x + (1/3x)²
= 9x² + 2 + 1/9x²
(viii) (2x – 1/2x)²
= (2x)² – 2.2x.1/2x + (1/2x)²
= 4x² + 1/4x² – 2
Question no – (2)
Solution :
(i) x + 3y
= x² + 2. x.3y + (3y)²
= x² + 6xy + 9y²
Hence, the square of x + 3y is x² + 6xy + 9y²
(ii) 2x – 5y
= (2x – 5y)²
= (2x)² – 2.2x.5y + (5y)²
= 4x² – 20y + 25y²
Thus, the square of 2x – 5y is 4x² – 20y + 25y².
(iii) a + 1/5a
= (a + 1/5)²
= (a)² + (1/5a)² + 2.a.1/5a
= a² + 1/25a² + 2/5
Thus, the square of a + 1/5a is a² + 1/25a² + 2/5.
(iv) 2a – 1/a
= (2a – 1/a)²
= (2a)² – 2.2a.1/a + 1/a²
= 4a² – 4 + 1/a²
Therefore, the square of 2a – 1/a is 4a² – 4 + 1/a².
(v) x – 2y + 1
= (x – 2y + 1)²
= (x)² + (-2y)² + (1) + 2 × x × (- 2y) + 2 × (- 2y) × 1 + 2 × 1 × x
= x² + 4y² + 1 – 4xy – 4y + 2x.
Hence, the square of x – 2y + 1 is x² + 4y² + 1 – 4xy – 4y + 2x.
(vi) 3a – 2b – 5c
= (3a – 2b – 5c)²
= (3a)² + (-2b)² + (-5c)² + 2. × 3a × (-2b) + 2 × (-2b) (-5c) + 2 × – 5c × 3a
= 9a² + 4b² + 25c² – 12ab + 20bc – 30ca
Thus, the square of 3a – 2b – 5c is 9a² + 4b² + 25c² – 12ab + 20bc – 30ca.
(vii) 2x + 1/x + 1
= (2x + 1/x + 1)²
= (2x)² + (1/x)² + 1² + 2 × 2x × 1/x + 2 × 1/x × 1 + 2 × 1 × 2x
= 4x² + 1/x² + 1 + 4 + 2/x + 4x
= 4x² + 1/x² + 5 + 2/x + 4x
Hence, the square of 2x + 1/x + 1 is 4x² + 1/x² + 5 + 2/x + 4x.
(viii) 5 – x + 2/x
= (5 – x + 2/x)²
= (5)² + (- x)² + (2/x)² + 2 × 5 × (- x) + 2 × (- x) × 2/x + 2 × 2/x + 5
= 25 + x² + 4/x² – 10x – 4 + 20/x
= 21 + x² + 4/x² – 10x + 20/x
Therefore, the square of 5 – x + 2/x is 21 + x² + 4/x² – 10x + 20/x.
(ix) 2x – 3y + z
= (3x – 3y + z) ²
= 4x² + 9y² + z² – 12xy – 6yz + 4zx
Thus, the square of 2x – 3y + z is 4x² + 9y² + z² – 12xy – 6yz + 4zx.
(x) x + 1/x – 1
= (x + 1/x – 1)²
= (x)² + (1/x)² – (1)² + 2 × x × 1/x + 2 × 1/x × (- 1) + 2 × (- 1) × x
= x² + 1/x² + 1 + 2 – 2/x – 2x
= x² + 1/x² + 3 – 2/x – 2x
Therefore, the square of x + 1/x – 1 is x² + 1/x² + 3 – 2/x – 2x.
Question no – (3)
Solution :
(i) (208)²
= (200 + 8)²
= (200)² + 2. 200. 8 + 64
= 40000 + 3200 + 64
= 43264
(ii) (92)²
= (100 – 8)²
= (100)² – 2. 100. 8 + 64
= 10000 – 1600 + 65
= 8464
(iii) (415)²
= (400 + 15)²
= (400)² + 2. 400. 15 + 225
= 160000 + 12000 + 225
= 172225
(iv) (188)²
= (200 – 12)²
= (200)² – 2. 200. 12 + (12)²
= 40000 – 4800 + 144
= 35344
(v) (9.4)²
= (10 – 0.6)²
= (10)² – 2. 10. 0.6 + (0.6)²
= 100 + 0.36 – 12
= 88.36
(vi) (20.7)²
= (20 + 0.7)²
= 400 + 0.49 + 2. 20 × 0.7
= 400 + 0.49 + 28
= 428.49
Question no – (4)
Solution :
(i) (2a + b)³
= (2a)³ + (b)³ + 3 × 2a × b (2a + b)
= 8a³ + b³ + 12a²b + 6ab²
(ii) (a – 2b)³
= (a)³ – (2b)³ – 3 × a × 2b (a – 2b)
= a³ – 8b³ – 6ab (a – 2b)
= a³ – 8b³ – 6a²b + 12ab²
(iii) (3x – 2y)³
= (3x)³ – (2y)³ – 3 × 3x × 2y (3x – 2y)
= 27x³ – 8y³ – 18xy (3x – 2y)
= 27x³ – 8y³ – 54x²y + 36xy²
(iv) (x + 5y)³
= (x)³ + (5y)³ + 3. x. 5y (x + 5y)
= x³ + 12xy³ + 15xy (x + 5y)
= x³ + 12xy³ + 15x²y + 75xy²
(v) (a + 1/a)³
= a³ + 1/a³ + 3. a. 1/a (a + 1/a)
= a³ + 1/a³ + 3 (a + 1/a)
= a³ + 1/a³ + 3a + 3/a
(vi) (2a – 1/2a)³
= (2a)³ – (1/2a)³ – 3 × 2a × 1/2a (2a – 1/2a)
= 8a³ – 1/8a³ – 3 (2a – 1/2a)
= 8a³ – 1/8a³ – 6a + 3/2a
Question no – (5)
Solution :
(i) a + 2
∴ (a + 2)³
= a³ + 2 + 3 × a × 2a (a + 2)
= a³ + 8 + 6a² + 12a
Therefore, the cube of a + 2 is a³ + 8 + 6a² + 12a.
(ii) 2a – 1
∴ (2a – 1)³
= (2a)³ – 1³ – 3 × 2a × 1 (2a – 1)
= 8a³ – 1 – 12a² + 6a
= 8a³ – 12a² + 6a – 1
Therefore, the cube of 2a – 1 is 8a³ – 12a² + 6a – 1.
(iii) 2a + 3b
∴ (2a + 3b)³
= (2a)³ + (3b)³ + 3 × 2a × 3b (2a + 3b)
= 8a³ + 27b³ + 18ab (2a + 3b)
= 8a³ + 27b³ + 36acb + 54ab²
Therefore, the cube of 2a + 3b is 8a³ + 27b³ + 36acb + 54ab²
(iv) 3b – 2a
∴ (3b – 2a)³
= (3b)³ – (2a)³ – 3 × 3b × 2a (3b – 2a)
= 27b³ – 8a³ – 18ab (3b – 2a)
= 27b³ – 8a³ – 54ab² + 36a²b
Hence, the cube of 3b – 2a is 27b³ – 8a³ – 54ab² + 36a²b.
(v) 2x + 1/x
∴ (2x + 1/x)³
= (2x)³ + (1/x)³ + 3 × 2x × 1/x (2x + 1/x)
= 8x³ + 1/x³ + 6 (2x + 1/x)
= 8x³ + 1/x³ + 12x + 6/x
Thus, the cube of 2x + 1/x is 8x³ + 1/x³ + 12x + 6/x.
(vi) x – 1/2
∴ (x – 1/2) ³
= (x) ³ – (1/2) ³ – 3 × x × 1/2 (x – 1/2)
= x³ – 1/8 – 3x/2 (x – 1/2)
= x³ – 1/8 + 3x³ /1 + 3x/4
Therefore, the cube of x – 1/2 is x³ – 1/8 + 3x³ /1 + 3x/4
Identities Exercise 12(C) Solution :
Question no – (1)
Solution :
As per the question,
a + b = 5 and ab = 6,
∴ a² + b²
= (a + b)² – 2ab
= (5)² – 2 × 6
= 25 – 12
= 13
Therefore, a² + b² will be 12
Question no – (2)
Solution :
According to the question, a – b = 6 and ab = 16
∴ a² + b²
= (a + b)² + 4ab
= (6)² + 4 × 16
= 36 + 64
= 100
Therefore, a² + b² will be 100
Question no – (3)
Solution :
As per the question we know,
a² + b² = 29
ab = 10
(i) ∴ a + b
= (a + b)² – 2ab = 29
= (a + b)² = 29 + 2 × 10
= 29 + 20
= 49
= a + b = √49
= a + b = 7
Hence, a + b = 7
(ii) ∴ a – b
= (a – b)²
= a² – 2ab + b²
= a² + b² – 2ab
= (a – b)²
= 29 – 2 × 10
= 29 – 20 = 9
∴ (a – b) = √9
= (a – b) = 3
Therefore, (a – b) = 3
Question no – (4)
Solution :
From the question we know,
a² + b² = 10,
ab = 3
(i) ∴ (a – b)²
= a² + b² – 2ab
= 10 – 2 × 3
= 10 – 6 = 4
= (a – b) = √4
∴ (a – b) = 2
Hence, (a – b) = 2
(ii) ∴ a + b
= (a + b)²
= a² + 2ab + b²
= a² + b² + 2ab
= 10 + 2 × 3
= 10 + 6
= 16
= (a + b) = √16
∴ (a + b) = 4
Therefore, (a + b) will be 4
Question no – (5)
Solution :
Given in the question,
a + 1/a = 3;
a² + 1/a² = ?
∴ (a + 1/a)²
= a² + 1/a² + 2
= (3)²
= 9
Now, a² + 1/a² + 2
= a² + 1/a²
= 9 – 2 = 7
Thus, a² + 1/a² will be 7
Question no – (6)
Solution :
As per the question,
a – 1/a = 4,
a² + 1/a² = ?
∴ (a – 1/a)²
= a² + 1/a² – 2
= (4)²
Now, a² + 1/a² – 2
= a² + 1/a²
= 16 + 2
= 18
Therefore, a² + 1/a² will be 18.
Question no – (7)
Solution :
According to the question,
a² + 1/a² = 23,
a + 1/a = ?
∴ (a + 1/a)²
= a² + 1/a² + 2. a. 1/a
= 23 + 2
Now, (a + 1/a)²
= √25
= 5
Thus, a + 1/a will be 5.
Question no – (8)
Solution :
Given, a² + 1/a² = 11,
a – 1/a = ?
∴ (a – 1/a)²
= a² + 1/a² – 2
= 11 – 2 = 9
Now, a – 1/a
= √9
= 3
Therefore, a – 1/a will be 3
Question no – (9)
Solution :
As per the question we know,
a + b + c = 10
a² + b² + c² = 38,
ab + bc + ca = ?
∴ (a + b+ c) = 10
= (a + b + c) ² = (10)
= a² + b² + c² + 2ab + 2bc + 2ca = 100
= 38 + 2 (ab + bc + ca) = 100
= 2 (ab + bc + ca)
= 100 – 38
= 62
Now, ab + bc + ca
= 62/2
= 31
Therefore, ab + bc + ca will be 31.
Question no – (10)
Solution :
Here, a + b + c = 9
= (a + b + c)² = (9)²
= a² + b² + c² + 2ab + 2bc + 2ca
= 81
= a² + b² + c² + 2 (ab + bc + ca)
= 81
= a² + b² + c² + 2 × 24
= 81
= a² + b² + c²
= 81 – 48
= 33
Hence, a² + b² + c² will be 33
Question no – (11)
Solution :
According to the question,
= (a + b + c) ² = a² + b² + c² + 2ab + 2bc + 2ca
= 83 + 2 (ab + bc + ca)
= 83 + 2 × 71
= 83 + 142
= 225
Now, (a + b + c)
= √225
= 15
Therefore, a + b + c = 15
Question no – (12)
Solution :
Given in the question,
a + b = 6
ab = 8
∴ a + b = 6
= (a + b) ³ = 6³
= 216
= a³ + b³ + 3ab (a + b) = 216
= a + b + 3 × 8 × 6 = 216
= a³ + b³ + 144
= 216
Now, a³ + b³
= 216 – 144
= 72
Thus, a³ + b³ will be 72.
Question no – (13)
Solution :
Here, a – b = 3
∴ (a – b)³ = 3 = 27
= a³ – b³ – 3ab(a-b) = 27
= a³ – b³ – 3 × 10 (3) = 27
= a³ – b³ – 90 = 27
= a³ – b³ = 27 + 90
= a³ – b³ = 117
Therefore, a³ – b³ will be 117.
Question no – (14)
Solution :
Given, a + 1/a = 5
∴ (a + 1/a)³ = 5³
= a³ + 1/a³ + 3.a.1/a (a + 1/a) = 125
= a³ + 1/a³ + 3 × 5 = 125
= a³ + 1/a³ = 125 – 15
= 110
Hence, a³ + 1/a³ will be 110
Question no – (15)
Solution :
Here, a – 1/a = 4
∴ (a – 1/a)³ = 4³ = 64
= a³ – 1/a³ – 3. a. 1/a (a – 1/a) = 64
= a³ – 1/a³ – 3 × 4 = 64
= a³ – 1/a³ = 64 + 12
= a³ – 1/a³ = 76
Thus, a³ – 1/a³ will be 76
Question no – (16)
Solution :
(i) Given, 4x² + 1/4x²
= (2x – 1/2x)² = 4² = 16
= (2x)² + (1/2x)² – 2. 2x. 1/2x = 16
= 4x² + 21/4x²
= 16 + 2
= 18
Therefore, 4x² + 1/4x² will be 18
(ii) 8x³ – 1/8x³
= 2x – 1/2x = 4
= (2x – 1/2x) = 4
= (2x) – (1/2x) – 3. 2x. 1/2x (2x – 1/2x) = 64
= 8x³ – 1/8x³ – 3 × 4 = 64
= 8x³ – 1/8x³ = 64 + 12
= 8x³ – 1/8x³ = 76
Thus, 8x³ – 1/8x³ will be 76.
Question no – (18)
Solution :
x² + y² = 13 ; xy = 6 (given)
(i) (x + y)² = x² + y² + 2xy
= 13 + 2 × 6
= 13 + 12
= 25
= x + y = ± 5
(ii) ( x -y)² = x² + y² – 2xy
= 13 – 2 × 6
= 13 – 12
= 1
∴ (x – y) = will be ± 1
Identities Exercise 12(D) Solution :
Question no – (1)
Solution :
(i) Given, (3x + 1/2) (2x + 1/3)
= 6x² + x + x + 1/6
= 6x² + 2x + 1/6
(ii) Given, (2a + 0.5) (7a – 0.3)
= 14a² – 0.6a + 3.5a – 0.15
= 14a² + 2.9a – 0.15
(iii) Given, 9 – y) (7 + y)
= 63 + 9y – 7y – y²
= 63 + 2y – y²
(iv) Given, (2 – z) (15 – z)
= 30 – 2z – 15z + z²
= z² – 17z + 30
(v) Given, (a² + 5) (a² – 3)
= a⁴ – 3a² + 5a² – 15
= a⁴ + 2a² – 15
(vi) Given, (4 – ab) (8 + ab)
= 32 + 4ab – 8ab – a²b²
= 32 – 4ab – a²b²
(vii) Given, (5xy – 7) (7xy + 9)
= 35x²y² + 45xy – 49xy – 63
= 35x²y² – 4xy – 63
(viii) Given, (3a² – 4b²) (8a² – 3b²)
= 24a⁴ – 9a²b² – 3²a²b² + 12b⁴
= 27a⁴ – 41a²b² + 12b⁴
Question no – (2)
Solution :
(i) Given, 2x – 3/5) (2x + 3/5)
= (2x)2 – (3/5)2
= 4x2 – 9/25
(ii) Given, (4/7a + 3/4b) (4/7a – 3/4b)
= (4/7a)2 – (3/4b)2
= 16/49a2 – 9/16b2
(iii) Given, (6 – 5xy) (6 + 5xy)
= (6)2 – (5xy)2
= 36 – 25x2y2)
(iv) Given, (2a + 1/2a) (2a – 1/2a)
= (2a)2 – (1/2a)2
= 4a2 – 1/4a2
(v) Given, (4x2 – 5y2) (4x2 + 5y2)
= (4x2)2 – (5y2)2
= 16x4 – 25y4
(vi) Given, (1.6x + 0.7y) (1.6x – 0.7y)
= (1.6x)2 – (0.7y)2
= 2.56x2 – 0.49y2
(viii) Given, (3x + 4y) (3x – 4y) (9x2 + 16y2)
= [(3x)2 – (4y)2] (9x2 + 16y2)
= (9x2 – 16y2) (9x2 + 16y2)
= (9x2)2 – (16y2)2
= 81x4 – 256y
(ix) Given, (a + bc) (a – bc) (a2 + b2c2)
= [ (a2 – (bc)2] (a2 + b2c2)
= (a2 – b2c2) (a2 + b2c2)
= (a2)2 – (b2c2)2
= a4 – b4c4
(x) 203 × 197
= (200 + 3) (200 – 3)
= (200)2 – (3)2
= 40000 – 9
= 39991
(xi) 20.8 × 19.2
= (20 + 0.8) (20 – 0.8)
= (20)2 – (0.8)2
= 400 – 0.64
= 399.36
Question no – (3)
Solution :
(i) 3x + 2/y
∴ (3x + 2/y)2
= (3x)2 + 2. 3x . 2/y + (2/y)2
= 9x2 + 4/y2 + 12x/y
Thus, the square of 3x + 2/y is 9x2 + 4/y2 + 12x/y.
(ii) 5a/6b – 6b/5a
∴ (5a/6b – 6b/5a)2
= (5a/6b)2 – (6b/5a)2
= (5a/6b)2 – 2 . 5a/6b × 6b/5a + (6b/5a)2
= 25a2/36b2 – 2 + 36b2/25a2
Hence, the square of 5a/6b – 6b/5a is 25a2/36b2 – 2 + 36b2/25a2
(iii) 2m2 – 2/3 n2
∴ (2m2 – 2/3 n2)2
= (2m2)2 – 2. 2m2 × 2/3 n2 + (2/3 n2)2
= 4m4 + 4/9 n4 – 8/3 m2 n2
= 4m4 – 8/3 m2n2 + 4/9 n2
So, the square of 2m2 – 2/3 n2 is 4m4 – 8/3 m2n2 + 4/9 n2
(iv) 5x + 1/5x
∴ (5x + 1/5x)2
= (5x)2 + 2. 5x . 1/5x + (1/5x)2
= 25x2 + 2 + 1/25x2
Thus, the square of 5x + 1/5x is 25x2 + 2 + 1/25x2
(v) 8x + 3/2y
∴ (8x + 3/2 y)2
= (8x)2 + 2 . 8x . 3y/2 + (3y/2)2
= 64x2 + 24xy + 9y2/4
Therefore, the square of 8x + 3/2y is 64x2 + 24xy + 9y2/4
(vi) 607
∴ (607)2
= (600 + 7)2
= (600)2 + 2 . 600 . 7 + 72
= 360000 + 8400 + 49
= 368449
Thus, the square of 607 is 368449.
(vii) 391
∴ (391)2
= (400 – 9)2
= (400)2 – 2 . 400 . 9 + 81
= 160000 – 7200 + 81
= 152881
Thus, the square of 391 is 152881.
(viii) 9.7
∴ (9.7)2
= (10 – 0.3)2
= (10)2 – 2 × 10 × 0.3 + (0.3)2
= 100 – 6 + 0.09
= 94.09
Therefore, the square of 9.7 is 94.09
Question no – (4)
Solution :
(i) a2 + 1/a2 ….[a + 1/a = 2 ]
= (a + 1/a)2 – 2
= (2)2 – 2
= 4 – 2
= 2
Therefore, a2 + 1/a2 will be 2
(ii) a4 + 1/a4
= (a2 + 1/a2)4 – 2
= (2)2 – 2
= 4 – 2
= 2
Hence, a4 + 1/a4 will be 2
Question no – (5)
Solution :
(i) m -1/m = 5 (given)
∴ m2 + 1/m2
= (m -1/m)2 + 2. m. 1/m
= (5)2 + 2
= 25 + 2
= 27
(ii) Given, m4 + 1/m4
= (m2 + 1/m2)2 – 2
= (27)2 – 2
= 729 – 2
= 727
(iii) m2 – 1/m2 = (m + 1/m) (m – 1/m)
= 5 (m + 1/m)
Now, (m + 1/m)2
= (m – 1/m)2 + 4
= 52 + 4
= 25 + 4
= 29
∴ m + 1/m = √29
Question no – (6)
Solution :
As per the question we know,
a2 + b2 = 41
ab = 4
(i) (a-b)2 = a2 – 2ab + b2
= 41 – 2 × 4
= 41 – 8 = 33
∴ a – b = √33
(ii) (a + b)2 = a2 + b2 + 2ab
= 41 + 2 × 4
= 41 + 8 = 49
∴ a + b = √49 = 7
Question no – (7)
Solution :
(i) 2a + 1/2a = 8
∴ 4a2 + 1/4a2
= (2a + 1/2a)2 – 2. 2a . 1/2a
= (8)2 – 2
= 64 – 2
= 62
(ii) 2a + 1/2a = 8
∴ 16a4 + 1/16a4
= (4a2 + 1/4a2)2 – 2 . 4a2 . 1/4a2
= (62)2 – 2
= 3844 – 2
= 3842
Question no – (8)
Solution :
According o the question,
3x – 1/3x = 5
(i) 9x2 + 1/9x2
= (3x – 1/3x)2 + 2 . 3x . 1/3x
= (5)2 + 2
= 25 + 2
= 27
(ii) 81x4 + 1/8x4
= (9x2 + 1/9x2)2 – 2
= (27)2 – 2
= 729 – 2
= 727
Question no – (9)
Solution :
(i) 3x – 4y + 5z)2
= (3x)2 + (-4y)2 + (5z)2 + 2 . (3x) . (-4y) + 2 (-4y) (5z) + 2 . (5z) (3x)
= 9x2 + 16y2 + 25z2 – 24xy – 40 yz + 30zx
(ii) 2a – 5b – 4c)2
= (2a)2 + (-5b)2 + (-4c)2 + 2 × 2a × (-5b) + 2 (-5b) (-4c) + 2 × (-4c) × (2a)
= 4a2 + 25b2 + 16c2 – 20ab + 40bc – 16ca
(iii) (5x + 3y)3
= (5x)3 + (3y)3 + 3 . 5x . 3y (5x + 3y)
= 125x3 + 27y3 + 225x2y + 135xy2
(iv) (6a – 7b)3
= (6a)3 – (7b)3 – 3 . 6a . 7b (6a – 7b)
= 216a3 – 343b3 – 126ab (6a – 7b)
= 216a3 – 343b3 – 756a2b + 882ab2
= 216a3 – 756a2b + 882ab2 – 343b3
Question no – (10)
Solution :
Given, a + b + c = 9
∴ (a + b + c)2 = a2 + b2 + c2 + 2 (ab + bc + ca)
= (9)2 = a2 + b2 + c2 + 2 × 15
= 81 = a2 + b2 + c2 + 30
= a2 + b2 + c2 = 81 – 30
= a2 + b2 + c2 = 51
Therefore, a2 + b2 + c2 = 51
Question no – (11)
Solution :
Given, (a + b + c) = 11
∴ (a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca)
= (11)2 = 81 + 2 (ab + bc + ca)
∴ 2 (ab + bc + ca) = 121 – 81 = 40
= (ab + bc + ca) = 40/2
= 20
Therefore, ab + bc + ca will be 20
Question no – (12)
Solution :
Given, 3x – 4y = 5 xy = 3
∴ 27x3 – 64y3
= (3x – 4y)3 + 3 . 3x . 4y (3x – 4y)
= (5)3 + 36 (xy) (3x – 4y)
= 125 + 36 (3) (5)
= 125 + 540
= 665
Therefore, 27x3 – 64y3 will be 665.
Question no – (13)
Solution :
a + b = 8, ab = 15 (given)
∴ a3 + b3
= (a + b)3 – 3ab (a + b)
= (8)3 – 3 × 15 × 8
= 512 – 360
= 152
Hence, a3 + b3 will be 152
Question no – (14)
Solution :
As per the given question,
3x + 2y = 9
xy = 3
Now, 27x3 + 8y3
= (3x)3 + (2y)3
= (3x + 2y)3 – 3 . 3x . 2y (3x + 2y)
= (9)3 – 18 × 3 × 9
= 729 – 486
= 243
Question no – (15)
Solution :
According to the question,
5x – 4y = 7
xy = 8
∴ 125x3 – 64y3
= (5x)3 – (4y)3
= (5x – 4y)3 + 3 . 5x . 4y (5x – 4y)
= 73 + 60 × 8 × 7
= 343 + 3360
= 3703
Therefore, 125x3 – 64y3 will be 3707
Question no – (16)
Solution :
Let, a and b two numbers,
∴ a – b = 5 and ab = 14
∴ a3 – b3 = (a – b)3 + 3ab (a – b)
= 53 + 3 × 14 × 5
= 125 + 210
= 335
Therefore, the difference between their cubes will be 335.
Next Chapter Solution :
👉 Chapter 13 👈
