Question

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

Answer 1

Given is ${\left(\frac{4}{3}\mathrm{m}-\frac{3}{4}\mathrm{n}\right)}^2+2\mathrm{mn}=\frac{16}{9}{\mathrm{m}}^2+\frac{9}{16}{\mathrm{n}}^2$ $\dots \dots \dots \dots \dots \dots \dots \dots \dots ..\left(\mathrm{i}\right)$

Let us solve $\mathrm{L}.\mathrm{H}.\mathrm{S}$ separately.

Take $\mathrm{L}.\mathrm{H}.\mathrm{S}={\left(\frac{4}{3}\mathrm{m}-\frac{3}{4}\mathrm{n}\right)}^2+2\mathrm{mn}$

Using the Identity ${\mathbf{(}\mathbf{a}\mathbf{-}\mathbf{b}\mathbf{)}}^{\mathbf{2}}\mathbf{=}{\mathbf{a}}^{\mathbf{2}}\mathbf{+}{\mathbf{b}}^{\mathbf{2}}\mathbf{-}\mathbf{2}\mathit{a}\mathit{b}$, we have

$\mathrm{L}.\mathrm{H}.\mathrm{S}={\left(\frac{4}{3}\mathrm{m}\right)}^2+{\left(\frac{3}{4}\mathrm{n}\right)}^2-2\times \frac{4}{3}\mathrm{m}\times \frac{3}{4}\mathrm{n}+2\mathrm{mn}$

$=\frac{16}{9}{\mathrm{m}}^2+\frac{9}{16}{\mathrm{n}}^2-2\mathrm{mn}+2\mathrm{mn}$

$=\frac{16}{9}{\mathrm{m}}^2+\frac{9}{16}{\mathrm{n}}^2$ $.......................................\left(\mathrm{ii}\right)$

From $\mathrm{equation}\left(\mathrm{i}\right)\mathrm{and}\left(\mathrm{ii}\right)$, it is clear that $\mathrm{L}.\mathrm{H}.\mathrm{S}=\mathrm{R}.\mathrm{H}.\mathrm{S}$.

Hence, proved.