Find the value
Question:

Find the value

$\left[x^{2}+1 / x^{2}\right]-4[x+1 / x]+6$

Solution:

$=x^{2}+1 / x^{2}-4 x-4 / x+4+2$

$=x^{2}+1 / x^{2}+4+2-4 x-4 x$

$=\left(x^{2}\right)+(1 / x)^{2}+(-2)^{2}+2 \times x \times 1 / x+2 \times 1 / x \times(-2)+2(-2) x$

Using identity

$x^{2}+y^{2}+z^{2}+2 x y+2 y z+2 z x=(x+y+z)^{2}$

We get,

$=[x+1 / x+(-2)]^{2}$

$=[x+1 / x-2]^{2}$

$=[x+1 / x-2][x+1 / x-2]$

 

$\therefore\left[x^{2}+1 / x^{2}\right]-4[x+1 / x]+6=[x+1 / x-2][x+1 / x-2]$

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