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Factorise each of the following: (i) 8a^3+b^3+12a^b+6ab^2 (ii)

Question

Factorise each of the following:
(i) $8 a^{3} + b^{3} + 12 a^{b} + 6 a b^{2}$
(ii) $8 a^{3} - b^{3} - 12 a^{2} b + 6 a b^{2}$
(iii) $27 - 125 a^{3} - 135 a + 225 a^{2}$
(iv) $64 a^{3} - 27 b^{3} - 144 a^{2} b + 108 a b^{2}$
(v) $27 p^{3} - \frac{1}{216} - \frac{9}{2} p^{2} + \frac{1}{4} p$

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Elijah Discussed

Factorise each of the following:
(i)

8 a^{3} + b^{3} + 12 a^{b} + 6 a b^{2}

(ii)

8 a^{3} - b^{3} - 12 a^{2} b + 6 a b^{2}

(iii)

27 - 125 a^{3} - 135 a + 225 a^{2}

(iv)

64 a^{3} - 27 b^{3} - 144 a^{2} b + 108 a b^{2}

(v)

27 p^{3} - \frac{1}{216} - \frac{9}{2} p^{2} + \frac{1}{4} p

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Text solutionVerified

We know that

(a + b)^{3} = a^{3} + b^{3} + 3 a^{2} b + 3 a b^{2}

and

(a - b)^{3} = a^{3} + b^{3} - 3 a^{2} b - 3 a b^{2}

(i)

8 a^{3} + b^{3} + 12 a^{2} b + 6 a b^{2} = (2 a)^{3} + (b)^{3} + 3 (2 a)^{2} (b) + 3 (2 a) (b)^{2} = (2 a + b)^{3}

(ii)

8 a^{3} + b^{3} - 12 a^{2} b - 6 a b^{2} = (2 a)^{3} + (b)^{3} - 3 (2 a)^{2} (b) - 3 (2 a) (b)^{2} = (2 a - b)^{3}

(iii)

27 - 125 a^{3} - 135 a + 225 a^{2} = (3)^{3} + (- 5 a)^{3} - 3 (3)^{2} (- 5 a) - 3 (3) (- 5 a)^{2} = (3 - 5 a)^{3}

(iv)

64 a^{3} - 27 b^{3} - 144 a^{2} b + 108 a b^{2} = (4 a)^{3} + (- 3 b)^{3} + 3 (4 a)^{2} (- 3 b) + 3 (4 a) (- 3 b)^{2} = (4 a - 3 b)^{3}

(v)

27 p^{3} + \frac{1}{216} - \frac{9}{2} p^{2} + \frac{1}{4} p

= (3 p)^{3} (- \frac{1}{6})^{3} + 3 (3 p) (- \frac{1}{6})^{2} + 3 (3 p) (- \frac{1}{6})^{2}

= (3 p - \frac{1}{6})^{3}

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Question Text	Factorise each of the following: (i) $8 a^{3} + b^{3} + 12 a^{b} + 6 a b^{2}$ (ii) $8 a^{3} - b^{3} - 12 a^{2} b + 6 a b^{2}$ (iii) $27 - 125 a^{3} - 135 a + 225 a^{2}$ (iv) $64 a^{3} - 27 b^{3} - 144 a^{2} b + 108 a b^{2}$ (v) $27 p^{3} - \frac{1}{216} - \frac{9}{2} p^{2} + \frac{1}{4} p$
Answer Type	Text solution:1
Upvotes	151