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?
 A: 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 $(\gamma^3+r)^3=-q^3\gamma^3$.
A: It’s not clear why you have written $$\frac{\gamma^2-2\alpha\beta}{\gamma^2}=\frac{1-2\alpha\beta\gamma}{\gamma^2\gamma}$$ This would mean that $\gamma^3=1$, which is not true.
It should be obvious to you that your answer is incorrect as you have arrived at a quadratic polynomial, not a cubic.
HINT…If you want a straightforward way of doing this question, try the following:
First find a polynomial whose roots are $\alpha^2$, $\beta^2$ and $\gamma^2$ which you can do either by using Vieta’s formulas or by substituting $y=x^2$ i.e. $x=\pm\sqrt{y}$.
Either way, you will get $$y^3+2qy^2+q^2y-r^2=0$$
Now, as you already know, $$\frac{\alpha^2+\beta^2}{\gamma^2}=-\frac{2q}{\gamma^2}-1$$
So now substitute $$z=-\frac{2q}{y}-1\implies y=-\frac{2q}{z+1}$$ into the above polynomial in $y$ and the resulting polynomial in $z$ will have the roots $\frac{\alpha^2+\beta^2}{\gamma^2}$ etc.
I hope this helps.
A: HINT.-From Vieta we get $$\alpha^2+\beta^2+\gamma^2=-2q\\(\alpha\beta)^2+(\alpha\gamma)^2+(\beta\gamma)^2=q^2\\(\alpha\beta\gamma)^2=r^2$$ Now instead of $a,b,c$ we consider $a+1=\dfrac{-2q}{\alpha^2},b+1=\dfrac{-2q}{\beta^2}$ and
$c+1=\dfrac{-2q}{\gamma^2}$ so the equation in $a+1,b+1,c+1$ is
$$\left(X+\dfrac{2q}{\alpha^2}\right)\left(X+\dfrac{2q}{\beta^2}\right)\left(X+\dfrac{2q}{\gamma^2}\right)=0$$ which can be easily simplified to
$$r^2X^3+2q^3X^2-8q^3x+8q^3=0$$ or better
$$r^2X^3+2q^3(X-2)^2=0$$
What remains is to go from roots to roots minus $1$. We are done.
