Question

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

Answer 1

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