If one root of x² + px + q = 0 may be the square of the other, then p³ + q² + q =

The given equation is $x^2+px+q=0$

Since one root is sqaure of the other, suppose that the roots of the abvoe equation are $\alpha ,{\alpha }^2$

Hence, sum of the roots is $\alpha +{\alpha }^2=-p$ and the product of the roots is ${\alpha }^3=q$

from both of the above equations eliminate $\alpha$ to get the condition

Substitute $\alpha =q^{\frac{1}{3}}$ in the equation $\alpha +{\alpha }^2=-p$

$q^{\frac{1}{3}}+q^{\frac{2}{3}}=-p$

Cubing on both sides 

$q+q^2+3q\left(-p\right)=-p^3\Rightarrow p^3+q^2+q=\boxed{3pq}$