Question

Factorise: $a^{3}b-\frac{b}{64}$

Answer 1

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

Using the above identity, the equation $a^{3}b-\frac{b}{64}$ or $b(a^3-\frac{1}{64})$can be factorised as follows:

$b(a^3-\frac{1}{64})=b\left[\right(a{)}^3-{\left(\frac{1}{4}\right)}^3]=b(a-\frac{1}{4})\left[\right(a{)}^2+{\left(\frac{1}{4}\right)}^2+(a\times \frac{1}{4})]=b(a-\frac{1}{4})(a^2+\frac{1}{16}+\frac{a}{4})$

Hence, $(a^{3}b-\frac{b}{64})=b(a-\frac{1}{4})(a^2+\frac{1}{16}+\frac{a}{4})$