Question

If $\sin \theta +\cos \theta =p,{\sin }^3\theta +{\cos }^3\theta =q$, then $p(p^2-3)=$

• A. $q$
• B. $2q$
• C. $-q$
• D. $-2q$

Answer 1

The correct option is D $-2q$

Given, $\sin \theta +\cos \theta =p$, ${\sin }^3\theta +{\cos }^3\theta =q$

We know $a^3+b^3=(a+b)(a^2+b^2-ab)$

Therefore, ${\sin }^3\theta +{\cos }^3\theta =(\sin \theta +\cos \theta )({\sin }^2\theta +{\cos }^2\theta -\sin \theta \cos \theta )$

$q=p(1-\sin \theta \cos \theta )$
$\frac{q}{p}=1-\sin \theta \cos \theta$

$\because \sin \theta +\cos \theta =p$

$1+2\sin \theta \cos \theta =p^2$

$2\sin \theta \cos \theta =p^2-1$

$\sin \theta \cos \theta =\frac{p^2-1}{2}$

$\Rightarrow \frac{q}{p}=[1-\frac{(p^2-1)}{2}]$

$\Rightarrow 2q=p(2-p^2+1)$

$\Rightarrow 2q=p(3-p^2)$

$\Rightarrow p(p^2-3)=-2q$