The equation ${\mathrm{x}}^3 – 3\mathrm{qx} + 2\mathrm{r} = 0$ has a repeated root then

$\begin{array}{l}Let\alpha ,\alpha ,\beta arerootsof{x}^3-3qx+2r=0\\ \therefore \alpha +\alpha +\beta =0\Rightarrow 2\alpha +\beta =0\Rightarrow \beta =-2\alpha --\left(1\right)\\ \therefore \alpha \left(\alpha \right)\beta =-2\gamma \Rightarrow {\alpha }^2\beta =-2\gamma --\left(2\right)\\ \therefore \alpha \left(\alpha \right)+\alpha \beta +\beta \alpha =-3q--\left(3\right)\\ put\left(1\right)in\left(2\right)\Rightarrow {\alpha }^2+2\alpha (-2\alpha )=-3q\\ \Rightarrow -3{\alpha }^2=-3q\\ \Rightarrow {\alpha }^2=q\Rightarrow \alpha =q^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\\ from\left(2\right)\\ {\alpha }^2(-2\alpha )=-2\gamma \\ \Rightarrow {\alpha }^3=\gamma \\ \Rightarrow q^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=\gamma \\ \Rightarrow q^3={\gamma }^2\end{array}$