If $\sin heta+\cos heta=p$ and $\sin ^3 heta+\cos ^3 heta=q$, then $p\left(p^2-3 ight)$ is equal to

Question

If $\sin 	heta+\cos 	heta=p$ and $\sin ^3 	heta+\cos ^3 	heta=q$, then $p\left(p^2-3ight)$ is equal to

Aicharya · Accepted Answer

$\begin{aligned} & 	ext { Given, } \quad \sin 	heta+\cos 	heta=p \ & 	ext { and } \quad \sin ^3 	heta+\cos ^3 	heta=q \ & \Rightarrow \quad(\sin 	heta+\cos 	heta) \ & \quad\left(\sin ^2 	heta-\sin 	heta \cdot \cos 	heta+\cos ^2 	hetaight)=q\end{aligned}$ $ \Rightarrow \quad p(1-\sin 	heta \cdot \cos 	heta)=q $ [From Eq. (i) and $\sin ^2 	heta+\cos ^2 	heta=1$ ] $ \begin{aligned} \Rightarrow & 1-\sin 	heta \cdot \cos 	heta=\frac{q}{p} \ \Rightarrow \quad & \sin 	heta \cdot \cos 	heta=1-\frac{q}{p} \end{aligned} $ On squaring both sides of Eq. (i), we get $ \begin{aligned} & \sin ^2 	heta+\cos ^2 	heta+2 \sin 	heta \cdot \cos 	heta=p^2 \ & \Rightarrow \quad 1+2\left(1-\frac{q}{p}ight)=p^2 \quad 	ext { [from Eq. (iii)] } \ & \Rightarrow \quad p+2(p-q)=p^3 \ & \Rightarrow \quad 3 p-2 q=p^3 \ & \Rightarrow \quad p^3-3 p=-2 q \ & \Rightarrow \quad p\left(p^2-3ight)=-2 q \ & \end{aligned} $

If $\sin \theta+\cos \theta=p$ and $\sin ^3 \theta+\cos ^3 \theta=q$, then $p\left(p^2-3\right)$ is equal to

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