Question

using suitable identities factorise the following
${x}^{3} - \frac{1}{8} {y}^{3}$
​

Answer 1

$\sf x^3 - \dfrac{1}{8} y^3$

$\sf \implies (x)^3 - \Bigg (\dfrac{1}{2} y \Bigg )^3$

$\rm \implies \Bigg (x-\dfrac{1}{2}y \Bigg ) \Bigg [(x)^2+\Bigg (\dfrac{1}{2}y \Bigg )^2 + x \times \dfrac{1}{2}y \Bigg ] \qquad ...[ since, \ a^3-b^3=(a-b)(a^2+b^2+ab)]$

$\rm \implies \Bigg (x-\dfrac{1}{2}y \Bigg ) \Bigg (x^2+\dfrac{1}{4}y^2 + \dfrac{1}{2}xy \Bigg )$

Or

$\rm \implies \Bigg (x-\dfrac{1}{2}y \Bigg ) \Bigg (x^2+\dfrac{1}{4}y^2 + \dfrac{xy}{2} \Bigg )$

Hope it helps...