Question

If $a{\cos }^3\theta +3a\cos \theta {\sin }^2\theta =m$, $a{\sin }^3\theta +3a\sin \theta {\cos }^2\theta =n$ then prove that $(m+n{)}^{\frac{2}{3}}+(m-n{)}^{\frac{2}{3}}=2a^{\frac{2}{3}}$.

Answer 1

$\frac{m+n}{a}={(\cos \theta +\sin \theta )}^3$

$\frac{m-n}{a}={(\cos \theta -\sin \theta )}^3$

$\to {(m+n)}^{2/3}+{(m-n)}^{2/3}=a^{2/3}\left[2\right]=2a^{2/3}$