Question

8 Factorise the following.
(i) 8a³ + b³ +12a²b + 6ab²
(ii) 8a³ - b³ - 12a²b + 6ab?
(iii) 27 – 125a³ - 135a + 225a²?
(iv) 64a - 27b³ – 144a²b + 108ab²?
(v) 27p³ - 1/216 - 9/2p² + 1/4p
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Answer 1

a^3+b^3+3a^2b+3ab^2=(a+b)^3

(i) 8a³ + b³ +12a²b + 6ab²

ans----=(2a)^3+(b)^3+3×2a×2a×b+3×2a×b

×b

=(2a+b)^3

ii)8a³ - b³ - 12a²b + 6ab

ans----=(2a)^3-(b)^3-3×2a×2a×b

+3×2a×b×b

=(2a-b)^3

iii)27 – 125a³ - 135a + 225a²

ans----=(3)^3-(5a)^3-3×3×3×5a

+3×3×5a×5a

=(3-5a)^3

iv) 64a^3 - 27b³ – 144a²b + 108ab²

ans----=(4a)^3-(3b)^3-3×4a×4a×3b

+3×4a×3b×3b

=(4a-3b)^3

v)27p³ - 1/216 - 9/2p² + 1/4p

ans----=(3p)^3-(1/6)^3-3×3p×3p×1/6

+3×3p×1/6×1/6

=(3p-1/6)^3

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Answer 2

Answer:

We know that

(a + b)3 = a 3 + b 3 + 3a 2b + 3ab 2 and (a − b)3 = a 3 + b 3 − 3a 2b − 3ab 2

(i) 8a 3 + b 3 + 12a 2b + 6ab 2 = (2a)3 + (b)3 + 3(2a)2(b) + 3(2a)(b)2 = (2a + b)3

(ii) 8a 3 + b 3 − 12a 2b − 6ab 2 = (2a)3 + (b)3 − 3(2a)2(b) − 3(2a)(b)2 = (2a − b)3

(iii) 27 − 125a 3 − 135a + 225a 2 = (3)3 + ( − 5a)3 − 3(3)2( − 5a) − 3(3)( − 5a)2 = (3 − 5a)3

(iv) 64a 3 − 27b 3 − 144a 2b + 108ab 2 = (4a)3 + ( − 3b)3 + 3(4a)2( − 3b) + 3(4a)( − 3b)2 = (4a − 3b)3

(v) 27p 3 + 1/216 − 9 2p 2 + 1/ 4 p = (3p)3 − 1/ 6 3 + 3(3p) − 1/ 6 2 + 3(3p) − 1/ 6 2 = 3p − 1/ 6 3