Factorise each of the following:
(i) $ 8 a^{3}+b^{3}+12 a^{2} 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 $

To do:

We have to factorise each of the given expressions.

Solution:

We know that,

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

$(a-b)^3=a^3-b^3-3ab(a-b)$

Therefore,

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

$= (2a + b)^3$

Hence $8a^3 + b^3 + 12a^2b + 6ab^2=(2a + b)^3$.

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

$= (2a - b)^3$

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

(iii) $27-125a^3-135a+225a^2=(3)^3 - (5a)^3 - 3(3)(5a)(3-5a)$

$=(3-5 a)^3$

Hence $27-125a^3-135a+225a^2=(3-5 a)^3$

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

$=(4a-3b)^3$

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

(v) $27 p^{3}-\frac{1}{216}-\frac{9}{2} p^{2}+\frac{1}{4} p=(3 p)^{3}-(\frac{1}{6})^{3}-3(3 p)(\frac{1}{6})(3 p-\frac{1}{6})$

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

Hence $27 p^{3}-\frac{1}{216}-\frac{9}{2} p^{2}+\frac{1}{4} p=(3 p-\frac{1}{6})^{3}$.

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Updated on: 10-Oct-2022

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