Question

Factorise: ${(a+\frac{1}{a})}^2-4{(x-\frac{1}{x})}^2$

Answer 1

We know that $a^2-b^2=(a+b)(a-b)$

Using the above identity, the expression ${(a+\frac{1}{a})}^2-4{(x-\frac{1}{x})}^2$can be factorised as follows:

${(a+\frac{1}{a})}^2-4{(x-\frac{1}{x})}^2={(a+\frac{1}{a})}^2-{\left[2(x-\frac{1}{x})\right]}^2=[(a+\frac{1}{a})-2(x-\frac{1}{x})][(a+\frac{1}{a})+2(x-\frac{1}{x})]$

Hence, ${(a+\frac{1}{a})}^2-4{(x-\frac{1}{x})}^2=[(a+\frac{1}{a})-2(x-\frac{1}{x})][(a+\frac{1}{a})+2(x-\frac{1}{x})]$