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