Question

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

Answer 1

Factorise the given expression :

The given expression is

$$\begin{gathered} x^2+\frac{1}{x^2}-3 \\ =x^2+\frac{1}{x^2}-2-1 \\ =x^2+\frac{1}{x^2}-2\left(x\right)\left(\frac{1}{x}\right)-1 \\ ={\left(x-\frac{1}{x}\right)}^2-1^2 \end{gathered}$$

using identity $\left(a^2-b^2\right)=(a+b)(a-b)$

$=\left(x-\frac{1}{x}+1\right)\left(x-\frac{1}{x}-1\right)$

Hence the answer is $\left(x-\frac{1}{x}+1\right)\left(x-\frac{1}{x}-1\right)$