Question

Find the following squares by using the identity
${(\frac{2a}{3b}+\frac{3b}{2a})}^2$

Answer 1

${(\frac{2a}{3b}+\frac{3b}{2a})}^2$

We have the identity ${(x+y)}^2=x^2+y^2+2xy$

${(\frac{2a}{3b}+\frac{3b}{2a})}^2={\left(\frac{2a}{3b}\right)}^2+{\left(\frac{3b}{2a}\right)}^2+2\times \frac{2a}{3b}\times \frac{3b}{2a}=\frac{{4a}^2}{{9b}^2}+\frac{{9b}^2}{{4a}^2}+2$