Question

If $a{\cos }^3\alpha +3a\cos \alpha {\sin }^2\alpha =m$ and

$a{\sin }^3\alpha +3a{\cos }^2\alpha \sin \alpha =n$, Then $(m+n{)}^{\frac{2}{3}}+(m-n{)}^{\frac{2}{3}}$

is equal to

• A. $2a^2$
• B. $2a^{\frac{1}{3}}$
• C. $2a^{\frac{2}{3}}$
• D. $2a^3$

Answer 1

The correct option is C

$2a^{\frac{2}{3}}$

Adding and subtracting the given relation,

we get $(m+n)=a{\cos }^3\alpha +3a\cos \alpha {\sin }^2\alpha$

+ $3a{\cos }^2\alpha .\sin \alpha +a{\sin }^3\alpha$

= $a(\cos \alpha +\sin \alpha {)}^3$

and similarly $(m-n)=a(\cos \alpha -\sin \alpha {)}^3$

Thus, $(m+n{)}^{\frac{2}{3}}+(m-n{)}^{\frac{2}{3}}$
$=a^{\frac{2}{3}}$ {$(\cos \alpha +\sin \alpha {)}^2+(\cos \alpha -\sin \alpha {)}^2$}
$=a^{\frac{2}{3}}$ $\left[2\right({\cos }^2\alpha +{\sin }^2\alpha \left)\right]$ = $2a^{\frac{2}{3}}$.