Question
Show that:
$\big(\frac{4\text{m}}{3}−\frac{3\text{n}}{4}\big)^2+2\text{mn}=\frac{16\text{m}^2}{9}+\frac{9\text{n}^2}{16}$
$\big(\frac{4\text{m}}{3}−\frac{3\text{n}}{4}\big)^2+2\text{mn}=\frac{16\text{m}^2}{9}+\frac{9\text{n}^2}{16}$
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