Finding an equation with related roots $f(x)=x^3-8x+14$ has roots $\alpha$, $\beta$ and $\gamma$.
I know that to find an equation with roots $3\alpha-2$, $3\beta-2$ and $3\gamma-2$, I just evaluate $f(\frac{x+2}{3})$, and to find an equation with roots $\alpha^3$, $\beta^3$ and $\gamma^3$, I just evaluate $f(x^\frac{1}{3})$.
However, what if an equation has roots $\alpha^2+\beta^2$, $\beta^2+\gamma^2$ and $\alpha^2+\gamma^2$, or another has roots $\alpha\beta$, $\beta\gamma$ and $\alpha\gamma$? What do I evaluate then?
 A: An equation which has $\alpha\beta$, $\alpha\gamma$, and $\beta\gamma$ as roots is, of course $P(x)=0$, with$$P(x)=(x-\alpha\beta)(x-\alpha\gamma)(x-\beta\gamma).$$On the other hand, note that$$P(x)=x^3-(\alpha\beta+\alpha\gamma+\beta\gamma)x^2+(\alpha^2\beta\gamma+\alpha\beta^2\gamma+\alpha\beta\gamma^2)x-(\alpha\beta\gamma)^2.$$Since $\alpha$, $\beta$, and $\gamma$ are the roots of $x^3-8x+14$, you know that $\alpha+\beta+\gamma=0$, that $\alpha\beta+\alpha\gamma+\beta\gamma=-8$, and that $\alpha\beta\gamma=-14$. So,$$\alpha^2\beta\gamma+\alpha\beta^2\gamma+\alpha\beta\gamma^2=\alpha\beta\gamma(\alpha+\beta+\gamma)=0$$and therefore $P(x)=x^2+8x^2-196$.
Can you do the same thing in the other case?
A: Hint:
$$\alpha\beta=\dfrac{14}\gamma$$
So, we need to find $f\left(\dfrac{14}y\right)$
and if $y=\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta$
$=(-\gamma)^2-2\cdot14/\gamma$
$$\implies\gamma^3-y\gamma-28=0$$
and $$\gamma^3-8\gamma+14=0$$
Solve for $\gamma,\gamma^3$
and use $\gamma^3=(\gamma)^3$
A: We obtain $$\sum_{cyc}(\alpha^2+\beta^2)=0,$$ which gives
$$\alpha^2+\beta^2+\gamma^2=0$$.
In another hand $$-14=\prod_{cyc}(\alpha^2+\beta^2)=\prod_{cyc}(-\alpha^2)=-\alpha^2\beta^2\gamma^2$$ and
$$-8=\sum_{cyc}(\alpha^2+\beta^2)(\alpha^2+\gamma^2)=\sum_{cyc}(\alpha^4+3\alpha^2\beta^2)=\sum_{cyc}\alpha^2\beta^2.$$
Now, we see that $\prod\limits_{cyc}\alpha\beta=14$ and $\sum\limits_{cyc}(\alpha\beta)^2=-8$ and
from $\alpha^2+\beta^2+\gamma^2=0$ we obtain $$2\sum_{cyc}\alpha\beta=(\alpha+\beta+\gamma)^2.$$
Thus, it remains to find $$\sum_{cyc}\alpha\beta\alpha\gamma=\alpha\beta\gamma(\alpha+\beta+\gamma),$$
for which we need to find $\alpha+\beta+\gamma$, for which  we need to solve a quartic equation.
Finally, I think we can not get here something nice. 
