Question

Factorise:

$(x^2+\frac{1}{x^2})-4(x+\frac{1}{x})+6$

Answer 1

$(x^2+\frac{1}{x^2})-4(x+\frac{1}{x})+6\Rightarrow (x^2+\frac{1}{x^2}+2)-4(x+\frac{1}{x})+4\Rightarrow {(x+\frac{1}{x})}^2-2\times 2(x+\frac{1}{x})+(2{)}^2={(x+\frac{1}{x}-2)}^2\{\because a^2-2ab+b^2=(a-b{)}^2\}$