If one root of the equation $a{x}^2+bx+c=0$ be the square of the other, then $b^3+a{c}^2+a^{2}c=$

Let $\alpha ,{\alpha }^2$ are roots

$\alpha +{\alpha }^2=\frac{-b}{a}\to \left(1\right),\alpha .{\alpha }^2=\frac{c}{a}$

Cube on both sides  ${\alpha }^3=\frac{c}{a}$

${(\alpha +{\alpha }^2)}^3={\left(\frac{{\displaystyle -b}}{{\displaystyle a}}\right)}^3$

${\alpha }^3+{\alpha }^6+3\alpha {\alpha }^2(\alpha +{\alpha }^2)=\frac{-b^3}{a^3}$

${\alpha }^3+{\alpha }^6+3\cdot \frac{c}{a}\left(\frac{{\displaystyle -b}}{{\displaystyle a}}\right)=\frac{-b^3}{a^3}$

$\frac{c}{a}+\frac{c^2}{a^2}-3\frac{bc}{a^2}=\frac{-b^3}{a^3}$

$\frac{ac+c^2-3bc}{a^2}=\frac{-b^3}{a^3}$

$a(ac+c^2-3bc)=-b^3$

$a^{2}c+a{c}^2-3abc=-b^3$