If $α,β,γ$ are the roots of the equation $f(x)=x^3+qx+r=0$ then find the equation whose roots are, If $α,β,γ$ are the roots of the equation $f(x)=x^3+qx+r=0$ then find the equation whose roots are, $\frac{\beta^2+\gamma^2}{\alpha^2}$,$\frac{\alpha^2+\gamma^2}{\beta^2}$,$\frac{\beta^2+\alpha^2}{\gamma^2}$.
My solution goes like this:

We consider, $a=\frac{\alpha^2+\beta^2}{\gamma^2}$,$b=\frac{\beta^2+\gamma^2}{\alpha^2}$,$c=\frac{\gamma^2+\alpha^2}{\beta^2}$. Now, $$\alpha+\beta+\gamma=0,\alpha\beta+\beta\gamma+\gamma\alpha=q,\gamma\alpha\beta=-r$$ and hence,$\alpha^2+\beta^2+\gamma^2=-2q$. Also, $a=\frac{\alpha^2+\beta^2}{\gamma^2}=\frac{-2q-\gamma^2}{\gamma^2}=\frac{-2q}{\gamma^2}-1$ or $\gamma^2=\frac{-2q}{a+1}$. Also, $$a=\frac{\alpha^2+\beta^2}{\gamma^2}=\frac{\gamma^2-2\alpha\beta}{\gamma^2}=1-\frac{2\alpha\beta\gamma}{\gamma^2\gamma}=1+\frac{2r}{\frac{-2q}{a+1}\gamma}=1-\frac{r(a+1)}{q\gamma}$$ and hence,$\gamma=\frac{r(a+1)}{q(1-a)}$. Also, $a=\frac{\alpha^2+\beta^2}{\gamma^2}=\frac{\gamma^2-2\alpha\beta}{\gamma^2}=1-2\frac{\alpha\beta\gamma}{\gamma^3}=1+\frac{2r}{\gamma^3}$. Thus, $\gamma^3=\frac{2r}{a-1}$. Now, we have, $\gamma^3+q\gamma+r=0$. Thus, $\frac{2r}{a-1}+\frac{r(a+1)}{(1-a)}+r=0$, which implies $a^2-a-2=0$. Thus, the required equation is $x^2-x-2=0$.

Is the above solution correct? If not, then where is it going wrong? I dont get where is the mistake occuring?

To answer your question, rather than suggest an alternative method: the error is in "Thus, $\frac{2r}{a-1}+\frac{r(a+1)}{(1-a)}+r=0$, which implies $a^2-a-2=0$". In fact (since $a \ne 1$) it implies that $2r -r(a+1) +r(a-1) = 0$, which is identically true. Instead, use your (correct) expression for $\gamma^3$ and substitute it into $(\gamma^3+r)^3=-q^3\gamma^3$.
