Question

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

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Solution

## 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 ${\mathbf{\left(}\mathbf{a}\mathbf{-}\mathbf{b}\mathbf{\right)}}^{\mathbf{2}}\mathbf{=}{\mathbf{a}}^{\mathbf{2}}\mathbf{+}{\mathbf{b}}^{\mathbf{2}}\mathbf{-}\mathbf{2}\mathbit{a}\mathbit{b}$, we have$\mathrm{L}.\mathrm{H}.\mathrm{S}={\left(\frac{4}{3}\mathrm{m}\right)}^{2}+{\left(\frac{3}{4}\mathrm{n}\right)}^{2}-2×\frac{4}{3}\mathrm{m}×\frac{3}{4}\mathrm{n}+2\mathrm{mn}$ $=\frac{16}{9}{\mathrm{m}}^{2}+\frac{9}{16}{\mathrm{n}}^{2}-2\mathrm{mn}+2\mathrm{mn}$ $=\frac{16}{9}{\mathrm{m}}^{2}+\frac{9}{16}{\mathrm{n}}^{2}$ $.......................................\left(\mathrm{ii}\right)$ 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.

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