Question
Factorise:$\text{x}^2+\frac{1}{\text{x}^2}-3$
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Answer
$\text{x}^2+\frac{1}{\text{x}^2}-3$$=\text{x}^2+\frac{1}{\text{x}^2}-2-1$
$=\Big(\text{x}^2+\frac{1}{\text{x}^2}-2\Big)-1$
$=\bigg(\text{x}^2+\frac{1}{\text{x}^2}-2(\text{x}^2)\Big(\frac{1}{\text{x}^2}\Big)\bigg)-1^2$
$=\Big(\text{x}-\frac{1}{\text{x}}\Big)^2-1^2$
$=\Big(\text{x}-\frac{1}{\text{x}}-1\Big)\Big(\text{x}-\frac{1}{\text{x}}+1\Big)$
$=\Big(\text{x}^2+\frac{1}{\text{x}^2}-2\Big)-1$
$=\bigg(\text{x}^2+\frac{1}{\text{x}^2}-2(\text{x}^2)\Big(\frac{1}{\text{x}^2}\Big)\bigg)-1^2$
$=\Big(\text{x}-\frac{1}{\text{x}}\Big)^2-1^2$
$=\Big(\text{x}-\frac{1}{\text{x}}-1\Big)\Big(\text{x}-\frac{1}{\text{x}}+1\Big)$
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