<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] Hint: assume WLOG that the three roots $a,b,c$ are all distinct (the case of multiple roots follows by continuity, or could be handled separately). Then subtracting the equations $a^3+pa+r=0\,$ and $b^3+pb+r=0\,$ gives:
$$\require{cancel}
a^3-b^3 + p(a-b) = 0 \iff \cancel{(a-b)}\big(a^2+ab+b^2+p\big) = 0 \iff (a-b)^2=-p-3ab
$$
The problem then reduces to finding the equation with roots $-p-3ab\,$, $-p-3bc\,$, $-p-3ca\,$.
Let $x=-p-y\,$, then the equation in $y$ will have roots $3ab,3bc,3ca\,$. After calculating the symmetric functions, the equation is found to be $y^3-3py^2-27r^2=0\,$. Substituting back $y=-x-p$ then gives the equation in $x$ as $(x+p)^3+3p(x+p)^2+27r^2=0$.
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