Question

Expand: $(x-a)(x+a)(\frac{1}{x}-\frac{1}{a})(\frac{1}{x}+\frac{1}{a})$

Answer 1

Using the identity is $(x+y)(x-y)=x^2-y^2$.

Consider $(x-a)(x+a)(\frac{1}{x}-\frac{1}{a})(\frac{1}{x}+\frac{1}{a})$ we apply the above identity as follows:

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

Multiply the resulting expression as follows:

$(x^2-a^2)(\frac{1}{x^2}-\frac{1}{a^2})=(x^2\times \frac{1}{x^2})+(x^2\times -\frac{1}{a^2})+({-a}^2\times \frac{1}{x^2})+(-a^2\times -\frac{1}{a^2})$

$=1-\frac{x^2}{a^2}-\frac{a^2}{x^2}+1=2-\frac{x^2}{a^2}-\frac{a^2}{x^2}$

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