Question

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

Answer 1

$$\begin{gathered} x^2+\frac{1}{x^2}-3 \\ =x^2+\frac{1}{x^2}-2-1 \\ =\left[x^2+{\left(\frac{1}{x}\right)}^2-2\times x\times \frac{1}{x}\right]-1 \\ ={\left(x-\frac{1}{x}\right)}^2-1^2\left[a^2-2ab+b^2={\left(a-b\right)}^2\right] \\ =\left(x-\frac{1}{x}+1\right)\left(x-\frac{1}{x}-1\right)\left[a^2-b^2=\left(a-b\right)\left(a+b\right)\right] \end{gathered}$$