If one root is square of the other root of the equation $x^2+px+q=0$, the relation between $p$ and $q$ is 

Let the roots are $\alpha , {\alpha }^2.$

Given $\alpha +{\alpha }^2=-p,\alpha ,{\alpha }^2=q$

Now, $\alpha +{\alpha }^2=-p$

$\begin{array}{l}\Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\left(\alpha +{\alpha }^2\right)}^3=(-p{)}^3=-p^3\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\alpha }^3+{\alpha }^6+3{\alpha }^3\left(\alpha +{\alpha }^2\right)=-p^3\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}q+q^2-3pq=-p^3\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{p}^3-q(3p-1)+q^2=0\end{array}$