6.350 Form the equation of the third degree from its roots $x_{1}^{2}, x_{1} x_{2}$ and $x_{2}^{2}$, if $x_{1}$ and $x_{2}$ are the roots of the equation $x^{2}+p x+q=0$.
Solution. By the inverse Vieta's theorem, the desired equation has the form:

$y^{3}-\left(x_{1}^{2}+x_{1} x_{2}+x_{2}^{2}\right) y^{2}+\left(x_{1}^{3} x_{2}+x_{1} x_{2}^{3}+x_{1}^{2} x_{2}^{2}\right) y-x_{1}^{3} x_{2}^{3}=0$.

But by Vieta's theorem for the equation $x^{2}+p x+q=0$, we have:

$x_{1} x_{2}=q, x_{1}+x_{2}=-p$. Therefore, $x_{1}^{2}+x_{1} x_{2}+x_{2}^{2}=\left(x_{1}+x_{2}\right)^{2}-x_{1} x_{2}=p^{2}-q$, $x_{1}^{3} x_{2}+x_{1} x_{2}^{3}+x_{1}^{2} x_{2}^{2}=x_{1} x_{2}\left(x_{1}^{2}+x_{2}^{2}+x_{1} x_{2}\right)=q\left(p^{2}-q\right), x_{1}^{3} x_{2}^{3}=q^{3}$.

Answer: $y^{3}-\left(p^{2}-q\right) y^{2}+\left(p^{2} q-q^{2}\right) y-q^{3}=0$.