Question
Factorize:$\Big(\text{x}^2+\frac{1}{\text{x}^2}\Big)-4\Big(\text{x}+\frac{1}{\text{x}}\Big)+6$
✓
Answer
$\Big(\text{x}^2+\frac{1}{\text{x}^2}\Big)-4\Big(\text{x}+\frac{1}{\text{x}}\Big)+6$$=\text{x}^2+\frac{1}{\text{x}^2}-4\text{x}-\frac{4}{\text{x}}+4+2$
$=\text{x}^2+\frac{1}{\text{x}^2}+4+2-4\text{x}-4\text{x}$
$=\big(\text{x}^2\big)+\Big(\frac{1}{\text{x}}\Big)^2+(-2)^2+2\times\text{x}\times\frac{1}{\text{x}}+2\times\frac{1}{\text{x}}\times(-2)+2(-2)\text{x}$
Using identity
$x^2 + y^2 + z^2 + 2xy + 2yz + 2zx = (x + y + z)^2$
We get,
$=\Big[\text{x}+\frac{1}{\text{x}}+(-2)\Big]^2$
$=\Big[\text{x}+\frac{1}{\text{x}}-2\Big]^2$
$=\Big[\text{x}+\frac{1}{\text{x}}-2\Big]\Big[\text{x}+\frac{1}{\text{x}}-2\Big]$
$\therefore\Big[\text{x}^2+\frac{1}{\text{x}^2}\Big]-4\Big[\text{x}+\frac{1}{\text{x}}\Big]+6$
$=\Big[\text{x}+\frac{1}{\text{x}}-2\Big]\Big[\text{x}+\frac{1}{\text{x}}-2\Big]$
$=\text{x}^2+\frac{1}{\text{x}^2}+4+2-4\text{x}-4\text{x}$
$=\big(\text{x}^2\big)+\Big(\frac{1}{\text{x}}\Big)^2+(-2)^2+2\times\text{x}\times\frac{1}{\text{x}}+2\times\frac{1}{\text{x}}\times(-2)+2(-2)\text{x}$
Using identity
$x^2 + y^2 + z^2 + 2xy + 2yz + 2zx = (x + y + z)^2$
We get,
$=\Big[\text{x}+\frac{1}{\text{x}}+(-2)\Big]^2$
$=\Big[\text{x}+\frac{1}{\text{x}}-2\Big]^2$
$=\Big[\text{x}+\frac{1}{\text{x}}-2\Big]\Big[\text{x}+\frac{1}{\text{x}}-2\Big]$
$\therefore\Big[\text{x}^2+\frac{1}{\text{x}^2}\Big]-4\Big[\text{x}+\frac{1}{\text{x}}\Big]+6$
$=\Big[\text{x}+\frac{1}{\text{x}}-2\Big]\Big[\text{x}+\frac{1}{\text{x}}-2\Big]$
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