Question

Factorise:
$2a^3+\frac{1}{4}$

Answer 1

We know the identity $a^3+b^3=(a+b)(a^2+b^2-ab)$

Using the above identity, the equation $2a^3+\frac{1}{4}$ or $2(a^3+\frac{1}{8})$ can be factorised as follows:

$2(a^3+\frac{1}{8})=2\left[\right(a{)}^3+{\left(\frac{1}{2}\right)}^3]=2(a+\frac{1}{2})\left[\right(a{)}^2+{\left(\frac{1}{2}\right)}^2-(a\times \frac{1}{2})]=2(a+\frac{1}{2})(a^2+\frac{1}{4}-\frac{a}{2})$

Hence, $(2a^3+\frac{1}{4})=2(a+\frac{1}{2})(a^2+\frac{1}{4}-\frac{a}{2})$