Question

Prove that ${(\frac{4}{3}m-\frac{3}{4}n)}^2+2mn=\frac{16}{9}{m}^2+\frac{9}{16}{n}^2$.

Answer 1

Consider the LHS of the given equality as shown below:

${(\frac{4}{3}m-\frac{3}{4}n)}^2+2mn=[{\left(\frac{4}{3}m\right)}^2+{\left(\frac{3}{4}n\right)}^2-(2\times \frac{4}{3}m\times \frac{3}{4}n)]+2mn(\because (a-b{)}^2=a^2+b^2-2ab)=\frac{16}{9}{m}^2+\frac{9}{16}{n}^2-2mn+2mn=\frac{16}{9}{m}^2+\frac{9}{16}{n}^2$

Hence, ${(\frac{4}{3}m-\frac{3}{4}n)}^2+2mn=\frac{16}{9}{m}^2+\frac{9}{16}{n}^2$