Question

If $a{\cos }^3\theta +3a{\sin }^2\theta \cos \theta =m$ and $a{\sin }^3\theta +3a\sin \theta {\cos }^2\theta =n$, find $(m+n{)}^{2/3}+(m-n{)}^{2/3}=$

• A. $2a^{2/3}$.
• B. $2a^{4/3}$.
• C. $a^{2/3}$.
• D. $Noneofthese$

Answer 1

The correct option is A $2a^{2/3}$.

We have $m=a{\cos }^3\theta +3a{\sin }^2\theta \cos \theta$

& $n=a{\sin }^3\theta +3a\sin \theta {\cos }^2\theta$

Now, $(m+n{)}^{2/3}+(m-n{)}^{2/3}$

$={\left[\right(a{\cos }^3\theta +3a{\sin }^2\theta \cos \theta )+(a{\sin }^3\theta +3a\sin \theta {\cos }^2\theta \left)\right]}^{2/3}+{\left[\right(a{\cos }^3\theta +3a{\sin }^2\theta \cos \theta )-(a{\sin }^3\theta +3a\sin \theta {\cos }^2\theta \left)\right]}^{2/3}$

$=a^{2/3}{\left[\right({\sin }^3\theta +{\cos }^3\theta )+3a\sin \theta \cos \theta (\sin \theta +\cos \theta \left)\right]}^{2/3}+a^{2/3}{\left[\right({\cos }^3\theta -{\sin }^3\theta )-3a\sin \theta \cos \theta (\cos \theta -\sin \theta \left)\right]}^{2/3}$

[Now as we know, $a^3+b^3+3ab(a+b)=(a+b{)}^3$ & $(a^3-b^3-3ab(a-b)=(a-b{)}^3$]

So, $=a^{2/3}\left\{\right[(\cos \theta +\sin \theta {)}^3{]}^{2/3}+[(\cos \theta -\sin \theta {)}^3{]}^{2/3}\}$

$=a^{2/3}\left[\right(\cos \theta +\sin \theta {)}^2+(\cos \theta -\sin \theta {)}^2]$

$=a^{2/3}({\cos }^2\theta +2\sin \theta \cos \theta +{\sin }^2\theta +{\cos }^2\theta -2\sin \theta \cos \theta +{\sin }^2\theta )$

$=a^{2/3}(1+1)$

$=2a^{2/3}$

So, $(m+n{)}^{2/3}+(m-n{)}^{2/3}=2a^{2/3}$