Question
Factorise $\text{x}^2+\frac{1}{\text{x}^2}+2-3\text{x}-\frac{3}{\text{x}}$.
✓
Answer
$\text{x}^2+\frac{1}{\text{x}^2}+2-3\text{x}-\frac{3}{\text{x}}$ (given)$=\text{x}^2+\frac{1}{\text{x}^2}+2.\text{x}.\frac{1}{\text{x}}-3\Big(\text{x}+\frac{1}{\text{x}}\Big)$
$=\Big(\text{x}+\frac{1}{\text{x}}\Big)^2-3\Big(\text{x}+\frac{1}{\text{x}}\Big)$
$=\Big(\text{x}+\frac{1}{\text{x}}\Big)\Big(\text{x}+\frac{1}{\text{x}}-3\Big) (\text{taking}\Big(\text{x}+\frac{1}{\text{x}}\Big)\text{is common})$
$=\Big(\text{x}+\frac{1}{\text{x}}\Big)^2-3\Big(\text{x}+\frac{1}{\text{x}}\Big)$
$=\Big(\text{x}+\frac{1}{\text{x}}\Big)\Big(\text{x}+\frac{1}{\text{x}}-3\Big) (\text{taking}\Big(\text{x}+\frac{1}{\text{x}}\Big)\text{is common})$
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