<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.-$$a+b+c=0\\ab+ac+bc=p\\abc=r$$ This gives 
$$(a-b)^2+(b-c)^2+(c-a)^2=(c^2+4bc+4b^2)+(a^2+4ac+4c^2)+(b^2+4ab+4b^2)
=5(a^2+b^2+c^2)+4p$$
Since $a^2+b^2+c^2+2(ab+ac+bc)=0$ we have $$(a-b)^2+(b-c)^2+(c-a)^2=-10p+4p=-6p$$ Let $A$ and $B$ the other Vieta coefficients we need so$$X^3+6pX^2+AX+B=0$$ Taking into account what we desire, we have
$$(a-b)^6+6p(a-b)^4+A(a-b)^2+C=0\\(b-c)^6+6p(b-c)^4+A(b-c)^2+C=0$$ This a linear system in $A$ and $B$ whose solution is
$$A=\frac{(a-b)^6+6p(a-b)^4-(b-c)^6-6p(b-c)^4}{(a-b)^2-(b-c)^2}\\C=\frac{(b-c)^2[(a-b)^6+6p((a-b)^4]-(a-b)^2[(b-c)^6+6p(b-c)^4]}{(a-b)^2-(b-c)^2}$$ What remains to be simplified.The denominator is equal to $(a-c)(a-2b+c)$. The two numerators are polynomials of degree six for $A$ and degree eight for $C$.
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