# NCTB Class 8 Math Chapter Four Exercise 4.2 Solution

NCTB Class 8 Math Chapter Four Exercise 4.2 Solution by Math Expert. Bangladesh Board Class 8 Math Solution Chapter 4 Algebraic Formulae and Applications Exercise 4.2 Solution.

 Board NCTB Class 8 Subject Mathematics Chapter 4 Chapter Name Algebraic Formulae and Applications Exercise 4.2 Solution

Exercise:4.2

1> Find the cube of the following expressions with the help of the formula :

(a) 3x+y

= (3x)3 +y3+3.3x.y.(3x+y)

= 27x3+ y3 + 9.xy (3x+y)

= 27x3 + y3 + 27x2y + 9xy2

(b) (x2+y)2

=(x2)3 + y3 + 3.x2.y (x2+y2)

= x6 + y3+3x2y (x2+y)

= x6+xy3 + 3x4y + 3x2y2

(c) (5p+2q)

= (5p)3 + (2q)3 + 3.5p.2q.(5p+2q)

= 125p3+ 8q3 + 30pq (5p+2q)

= 125p3+8q3+150p2q+ 60pq2

(e) (6p-7)3

= (6p)3 – (7)3 – 3.6p.7 (6p.7)

= 216p3 – 343 – 126p (6p-7)

(f) (ax-by)3

=(ax)3 – (by)3 – 3.ax.by (ax-by)

= a3x3 – b2y2 – 3a2x2by + 3axb2y2

(g) (2p2-3r2)3

= (2p)3 – (3r2)3 – 3.2p2.3r2.(2p2 – 3r2)

= 8p3– 27r3 – 18p2r2 (2p2 – 3r2)

= 8p3 – 27r3 – 36p4r2 + 54p2r4

(h) (x3 + 2)3

= (x3)3 + 23 + 3.x3.2 (x3 +2)

= x27+8+6x­3 (x3+2)

= x27 +8 +6x6+12x

(h) (a2b2– c2d2)3

= (a2b2)3-(c2d2)3-3.a2b2.c2d2 (a2b2-c2d2)

= a6b6-c6d6-3a2b2c2d2(a2b2-c2d2)

= a6b6-c6d6-3a4b4c2d2+3a2b2c4d4

(l) (a2b-b3c)3

= (a2b)3 – (b3c)3– 3.a2b.b3c (a2b-b3c)

= a6b- b9c-3a2b4c(a2b-b3c)

= a6b – b9c – 3a4b5c + 3a2b7c

(m) (x3-2y3)3

= (x3)3 – (2y3)3 – 3.x3.2y3.(x3-2y3)

= x27 – 8y9 – 6x3y3 (x3– 2y3)

= x27-8y9-6x6y3 + 12x3y6

(n) (11a – 12b)3

= (11a)3 – (12)3

= 1331a3-1728b3-396ab (11a-12b)

= 1331a3– 1728b3-4356a2b – 4752ab2

(o) (x3+y3)3

= (x3)3+(y3)3 + 3.x3y3(x3+y3)

= x9+y9+3x6y3 + 3x3y6

2> Simplicity:

(a) (3x+y)3+3(3x+y)2(3x-y)+3(3x+y)(3x-y)2+(3x-y)3

= (3x+y+3x-y)3

= (6x)3

= 216x3

(b) (2p+5q+5q-2p)3

= (10q)3

= 100q3

(c) (x+2y)3-3(x+2y)2(x-2y) + 3(x+2y)(x-2y)2-(x-2y)3

= (x+2y-x+2y)3

= (4y)3

= 64y3

(d) (6m+2)3– 3(6m+2)2(6m-4)+3.(6m+2)(6m-4)2-(6m-4)+3

= (6m+2-6m+4)3

=(6)3

= 216

(e) (x-y)3+ (x+y)3+6x (x2-y2)

= (x-y)3 + (x+y)3 + 3.2x (x+y) (x-y)

= (x-y)3 + (x+y)3 + 3{(x+y)(x-y)} (x+y+x-y)

= (x-y+x+y)3

= (2x)3

= 8x3

3> If a+b=8 and ab=15, what is the value of a3+b3 ?

A3+b3

= (a+b)3-3.ab(a+b)

= (8)3-3.15  [8]

= 512-360

=152

5> If 2x+3y=13 and xy=6, find the Value of 8x3+27y3

=  2x +3y=13 ; xy=6 (given)

8x3+27y3

= (2x)3 + (3y)3

= (2x + 3y)3 – 3(2x+3y)(2x+3y)

= (13)3 – 18.6.13

=793

6> If p-q=5, pq=3 , find the value of p3-q3

p-q=5, pq=3 (given)

p3-q3

= (p-q)3 +3.pq.(p-q)

= (5)3+3.3.5

=125+45

= 170

7> if x-2y=3, find the value of x3-8y3-18y

x-2y=3 (given)

x3-8y3-18xy

= (x)3-(2y)3-3xy(x-2y)

= (x)3 –(2y)3-3.x.2y [3]

= (x-2y)3

=(3)3

=27

8> If 4x-3=5 , prove that 64x3-27-180x=125

L.H.S  64x3-27-180x

= (4x)3-(3)3-3.4x.3 (4x-3)

= (4x)3-(3)3-3.4x.3 [5]

= (4x-3)3

= 53

=125

R.H.S  (proved)

9> If a=-3 and b=3, find the value of 8a3+36a2b+54ab2+27b3Given a=-3 and b=2

Therefore, 8a3 + 36a2b+54ab2+27b3

= (2a)3+(3b)3+3.(2a)2.3b + 3.2a.(3b)2

= (2a+3b)3

= { 2(-3)+3.2}3

=(-6+6)3

= 0

10> If a=7, find the value of a3+6a2+12a+1

A=7 [given]

A3+6a2+12a+1

= (a)3 + 3.a2.2d+3.a.(2)2+(2)3+1-23

= (a+2)3+1-8

= (7+2)3 + 1-8

= (9)3+1-8

= 729+1-8

= 730-8

= 722

11> if x=5, what is the value of x3-12x2+48x-64

X=5 [given]

X3+12x2+48x-64

= x3-3.x2.4+3.x.42-43

= (x-4)3

=(5-4)3

=1

12> If a2+b2=c2, prove that a6+b6+3a2b2c2=c6

a2+b2=c2 [given]

R.H.S  C6

= (A2+B2)3

= (a2)3 + (b2)3+ 3.a2b2(a2+b2)

= a6+b6+3a2b2c2

L.H.S (PROVED)

13> If x+1/x=4 . prove that x3+1/x3=52

X+1/x=4 (given)

X3 + 1/x3

= (x)3 + (1/x)3

= (x+1/x)3-3.x.1/x.(x+1/x)

=(4)3-3.4

= 64-12

= 52

R.H.S (PROVED)

14> If a-1/a=5, what is the value of a3-1/a3 ?

a-1/a=5 [given]

a3-1/a3

= (a-1/a)3+3.a.1/a(a-1/a)

= (5)3+3.5

= 125+15

=140

15> Find the product with the help of formula :

(a) (a2+b2)(a2-a2b2+b4)

= (a2)3 + (b2)3

= a6+b6

(b) (ax-by)(a2x2+abxy+b2y2)

= (ax)3– (by)3=

= a3x3 – b3y3

(c) (2ab2 – 1)(4a2b4+2ab2+1)

= (2ab2)3 – (1)3

= 8a3b6-1

(e) (7a+4b)(49a2-28ab+16b2)

= (7a)3 + (4b)

= 343a3+64b3

(f) (2a-1) (4a2+2a+1) (8a3+1)

= (2a)3 – (1)3 (2a)3+(1)3

= (8a3-1) (8a3+ 1)

= (8a3)2 – (1)2

= 64a6 – 1

(g) (x+a) (x2-ax+a2) (x-a)(x2+ax+a2)

= (x3+a3) (x3-a3)

= (x3)2– (a3)2

= x6-a6

(h) (5a+3b) (25a2 – 15ab + 9b2) (125a3 – 27b3)

= (5a)3+(3b)3 (5a)3-(3b)3

= (5a3+3b3) (5a3-3b3)

= (5a3)2 – (3b3)2

= 125a6– 27b6

Updated: March 23, 2021 — 3:03 pm