<s>[INST] Let $a,b,c$ be roots of $x^3+px+r=0$. Find the cubic whose roots are $(a-b)^2$,$ (b-c)^2$ and $(c-a)^2$ Question
Let $a,b,c$ be roots of $x^3+px+r=0$. Then find the cubic whose roots are $(a-b)^2, (b-c)^2$ and $ (c-a)^2$
Attempt
I have tried using Vieta's formulas to compute coefficients of the sought cubic.
For sum of roots we have $$\sum_{cyc}(a-b)^2 = 2\left(\sum_{cyc} a^2-\sum_{cyc}{ab}\right)\\
= 2\left(\sum_{cyc} a\right)^2-6\left(\sum_{cyc} ab\right)\\
= -6p$$
This is coefficient of $x^2$ in the sought cubic.
But now computing coefficient of $x^2$ requires simplifying factors like $(a-b)^2\cdot(b-c)^2$ which becomes very lengthy. 
Is there a shorter way around? 
Thanks!
 [/INST] $\sum_{cyc}(a-b)^2$ is a symmetric polynomial of degree two, so it must be a polynomial in the coefficients of $x^3+px+r$. The only possibilities to form a polynomial of degree $2$ are $(a+b+c)^2$ and $ab+bc+ca=p.$ Since all terms containing the factor $a+b+c$ must vanish, it can be only a multiple of $p$. From the special case $(a,b,c)=(-1,0,1)$, where $p=-1$ and our three roots are $4,1,1$, we obtain $$\sum_{cyc}(a-b)^2=-6p.$$ Similarly, $\sum_{cyc}(a-b)^2\cdot(b-c)^2$ can only be a multiple of $p^2,$ so from the special case mentioned above, we get
$$\sum_{cyc}(a-b)^2\cdot(b-c)^2=9p^2.$$ Finally, $(a-b)^2\cdot(b-c)^2\cdot(c-a)^2$ is the discriminant of the cubic polynomial, a known expression (en.wikipedia.org/wiki/Discriminant ), so $$(a-b)^2\cdot(b-c)^2\cdot(c-a)^2=-4p^3-27r^2.$$
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