Question
Factorise $: x^2+\frac{1}{x^2}-3$
✓
Answer
$ x^2+\frac{1}{x^2}-3$
$ =x^2+\frac{1}{x^2}-2 \times x \times \frac{1}{x}-1$
$ =\left(x-\frac{1}{x}\right)^2-1$
$ =\left(x-\frac{1}{x}\right)^2-(1)^2$
$ =\left(x-\frac{1}{x}-1\right)\left(x-\frac{1}{x}+1\right)\left[\because a^2-b^2=(a+b)(a-b)\right]$
$ =x^2+\frac{1}{x^2}-2 \times x \times \frac{1}{x}-1$
$ =\left(x-\frac{1}{x}\right)^2-1$
$ =\left(x-\frac{1}{x}\right)^2-(1)^2$
$ =\left(x-\frac{1}{x}-1\right)\left(x-\frac{1}{x}+1\right)\left[\because a^2-b^2=(a+b)(a-b)\right]$
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