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