Factorize:$ \left(x^{2}+\frac{1}{x^{2}}\right)-4\left(x+\frac{1}{x}\right)+6 $

Given :

\( \left(x^{2}+\frac{1}{x^{2}}\right)-4\left(x+\frac{1}{x}\right)+6 \)

To do :

We have to factorize the given expression.

Solution :

$(x^{2}+\frac{1}{x^{2}})-4(x+\frac{1}{x})+6 = (x^{2}+\frac{1}{x^{2}}+2)-4(x+\frac{1}{x})+4$

$=(x+\frac{1}{x})^{2}-2 \times 2(x+\frac{1}{x})+(2)^{2}$

$=(x+\frac{1}{x}-2)^{2}$

Hence, $(x^{2}+\frac{1}{x^{2}})-4(x+\frac{1}{x})+6 =(x+\frac{1}{x}-2)^{2}$.