West Bengal Board Class 8 Math Chapter 5 Solution

Chapter 5 Solution West Bengal Board : Class 8 Mathematics

West Bengal Board Class 8 Math Solution full Exercise 5.1, 5.2 and 5.3 by Experts. Here in this page WB Board Class 8 Student can find Ganit Prabha Class 8 Math Chapter 5 ঘনফল নির্ণয় Exercise 5.1, 5.2 and 5.3 Solution.

Board

WB Board
Class

8 (Eight)

Subject

Mathematics
Book Name

Ganit Prabha

Chapter

5


Exercise 5.1 Solution

Exercise 5.2 Solution

(iii) (x2y – z2)

= (x6y3 – 3x4y2z2 + 3x2yz4 – z6)

(iv) (l + b – 2c)3

= l3 + b3  – 8c3 + 3l2b + 3lb2 – 6cl2 – 12lbc – 6b2c + 12c2l + 12c2b

(v) (2.89)3 + (2.11)3 + 15 x 2.89 x 2.11

= 24.137 + 9.4 + 91. 46

= 125

(vi) (2m + 3n)3 + (2m – 3n)3 + 12n (4n2 – 9n2)

= (2m)3 + 3. (2m)2.3n + 3. 2m. (3n)2 + (3n)3 + (2m)3 – 3.(2m)2.3n + 3.2m.(3n)2 + (3n)3 + 48mn2 – 180mn2

= 8m3 + 36m2n + 54mn2 + 27n3 + 8m3 – 36m2n + 54mn2 + 27n3 + 48mn2 – 108mn2

∴ x3 + y3

= (x + y)3 – 3.xy. (x+y)

= (2)3 – 3.1(2)

= 8 – 6

= 2

(f) x =  y + z

=) x – y = z

=) (x – y)3 = (z)3

=) x3 – y3 – 3xy (x – y) = z3

=) x3 – y3 – 3xy. Z = z3

=) x3 – y3 – z3 -3xyz = 0

(k) m +n = 5 ; m n = 6

(m2 + n2) (m3 + n3)

= {(m +n)2 – 2mn} {(m+n)3 – 3mn (m + n)}

= {(5)2 – 2 x 6} {(5)3 – 3 x 6 x 5}

= (25 -12) (125 – 90)

= 13 x 35

= 455

Exercise 5.3 Solution

(1) (i) (x)3 + (9)3

= x3 + 729

(ii) (2a – 1) {(2a) + 2a.1 + 12}

= (2a – 1) (4a2 + 2a + 1)

(iii) (3 – 5c) {(3) + 3.5c + (5c)2}

= (3 – 5c) (9 + 15c + 15c2)

(iv) (a + b + c) (a + b)2 – (a + b) c + c2

=(a+b)3 + c3

(v) (2x – 1+ x + 1)3

= (3x)3

= 27x3

(vii) (4a – 5b) (16a2 + 2ab + 25b2)

= (4a)2 – (5b)3

= 64a3 – 125b3

(viii) a3b3 – c3d3

= (ab – cd) (a2b2 + abcd + c2d2)

(ix) (1 – 4y) {(1)2 + 1.4y + (4y)2}

= (1 – 4y) (1 + 4y + 16y2)

(x) (2n + 1) {(2n)2 + 2n.1 + 12}

= (2n + 1) (4n2 + 2n +1)

(2) (i) (a + b) (a – b) (a2 + ab + b2) (a2 – ab + b2)

= (a + b) (a2 – ab + b2) (a – b) (a2 + ab + b2)

= (a3 + b3) (a3 – b3)

= (a3)2 – (b3)2

= a6 – b6

(ii) (a – 2b) (a2 + 2ab + 4b2) (a3 + 8b3)

= (a3 – 8b3) (a3 + 8b3)

= (a3)2 – (87b3)2

= a6 – 64b6

(iii) (4a2 – 9) (4a2 – 6a + 9) (4a2 + 6a + 9)

= {(2a)2 – (3)2 (4a2 – 6a + 9) (4a2 + 6a +9)

= (2a + 3) (2a – 3) (4a2 – 6a + 9) (4a2 + 6a + 9)

= (2a + 3) (4a2 – 6a + 9) (2a – 3) (4a2 + 6a + 9)

= {(2a)3+ (3)3} {(2a)3 – (3)3}

= (8a3 + 27) (8a3 – 27)

= (8a3)2 – (27)2

= 64a6 – 729

(iv) (x – y) (x2 + xy + y2) + (y –z) (y2 +yz + z2) + (z – x) (z2 + zx + x2)

= (x3 – y3) + (y3 – z3) + (z3 – x3)

= x3 – y3 + y3 – z3 + z3 – x3

= 0

(v) (x + 1) (x2 – x + 1) + (2x – 1) (4x2 + 2x + 1) – (x – 1) (x2 + x 1)

= x3 + 13 + (2x)3 – (1)3 – (x)3 – (1)3

= x3 + 13 + 8x3 – 13 – x3 + 13

= 8x3 + 13

= 8x3 + 1

(6) (i) 1000a3 + 27b6

= (10)3 + (3b2)3

= (10a + 3b2) {(10a)2 – 10a.3b2 + (3b2)2}

= (10a +3b2) (100a2 – 30ab2 + 9b4)

(ii) 1 – 216z3

= (1)3 – (6z)3

= (1 – 6z) (12 + 1.6z + (6z)2)

= (1 – 6z) (1 + 6z +36z2)

(iii) m4 – m

= m (m3 – 1)

= m (m – 1) (m2 + m + 1)

(iv) 192a2 + 3

= 3 (64a3 + 1)

= 3 {(4a)3 + (1)3}

= 3 {(4a + 1) (4a)2 – 4a.1 + 1}

= 3 (4a + 1) (16a2 – 4a + 1)

(v) 16a4x3 + 54ay3

= 2a (8a3x3 + 27y3)

= 2a {(2ax)3 + (3y)3}

= 2a {(2ax + 3y) (2ax)2 – 2ax.3y + (3y)2}

= 2a (2a + 3y) (4a2x2 – 6axy + 9y2)

(vi) 729a3b3c3 – 125

= (9abc)3 – (5)3

= (9abc – 5) {(9abc)2 – 9abc.5 + 52}

= (9abc – 5) (81a2b2c2 – 45abc + 25)

(ix) x3 + 3x2y + 3xy2 + 2y3

= x3 + 3x2y + 3xy2 + y3 + y3

= (x+y)3 + y3

= (x+ y +y) {(x + y)2 – (x+y).y + y2}

= (x + 2y) {(x2 + 2xy + y2 – xy – y2 + y2}

= (x + 2y) (x2 + xy + y2)

(x) 1 + 9x + 27x2 + 28x3

= 1 + 9x + 27x2 + 27x3 + x3

= 13 + 3.13.3x + 3.1. (3x)2 + (3x)2 + x3

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

= (1+3x+x) {(1+3x)2 – (1+3x).x + x2}

= (1+4x) (12 + 2.1.3x + (3x)2 – x – 3x2 + x2)

= (1 + 4x) (1 + 6x + 9x2 – x – 2x2)

= (1 + 4x) (1 + 5x + 7x2)

(xi) x3 – 9y3 – 3xy (x + y)

= x3 – y3 – 8y3 – 3x2y + 3xy2

= (x3 – 3x2y + 3xy2 – y3) – (8y3)

= (x – y)3 – (2y)3

= {(x-y) – 2y} {(x-y) – (x –y_ .2y + (2y)2}

= (x – 3y) (x2 – 2xy + y2 + 2xy – 2y2 + 4y2)

= (x – 3y) (x2 – 3y2)

(xii) 8 – a3 + 3a2b + 3ab2 + b3

= 8 – (a3 – 3a2b + 3ab2 – b3)

= (2)3 – (a-b)3

= {2 – (a – b)} {(2)2 + 2.(a – b) + (a – b)2)}

= (2 – a + b) {(4 + 2a – 2b + a2 – 2ab + b2)

= (2 – a + b) (4 + 2a – 2b + a2 + b2 – 2ab)

(xiii) x6 + 3x4b4 + 3x4b4 + b6 + a3b3

= {(x2)3 + 3.(x2)2.b2 + 3x2 (b2)2 + (b2)3} + (ab)3

= (x2 + b2)3 + (ab)3

= {(x2 + b2) + ab} {(x2 + b2) – (x2 + b2) ab + (ab)2}

= (x2 + b2 + ab) {(x2)2 + 2x2b2 + (b2)2} – (x2 + b2) ab + (ab)2}

= (x2 + b2 + ab) (x4 + 2x2b2 + b4 – ab(x2 + b2) + a2b2)}
(xiv) x6 + 27

= (x2)3 + (3)3

= (x2 + 3) {(x2)2 – x2 + 3 + 32}

= (x2 + 3) (x4 – 3x2 + 9)

(xv) x6 – y6

= (x3)2 – (y3)2

= (x3 + y3) (x3 – y3)

= (x + y) (x2 – xy + y2) (x –y) (x2 + xy + y2)

(xvi) x12 – y12

= (x6)2 – (y6)2

= (x6 + y6) (x6 – y6)

= {(x2)3 + (y2)3} {(x3)2 – (y3)2}

= {(x2 + y2) (x2)2 – x2y2 + (y2)2} (x3 + y3) (x3 – y3)

= (x2 + y2) (x4 – x2y2 + y4) (x3 + y3 (x3 – y3)

(xviii) m3 + n3 – m(m2 – n2) + n(m – n)2

= (m – n) (m2 + mn + n2) – n (m – n) (m+ n) + n (m – n) (m- n)

= (m – n) (m2 + mn + n2) – m (m + n) + n (m – n)

= (m – n) (m2 + mn + n2 – m2 – mn + mn – n2)

= (m – n) mn

Updated: September 11, 2020 — 4:19 pm

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