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

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!

Answer 1

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 $(x+p)^3+3p(x+p)^2+27r^2=0$.

Answer 2

$\sum_{cyc}(a-b)^2$ 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 $$(a-b)^2\cdot(b-c)^2\cdot(c-a)^2=-4p^3-27r^2.$$

Answer 3

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 $C$.