Question

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

Answer 1

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

Using the above identity, the equation $a^3-b^3-a+b$ can be factorised as follows:

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

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

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