Form an equation whose roots are $(a-b)^2,(b-c)^2,(c-a)^2.$

Question

If $a,b,c$ are the roots of the equation $x^3+x+1=0,$ Then the equation whose roots are $(a-b)^2,(b-c)^2,(c-a)^2.$

Try: $a+b+c=0,ab+bc+ca=1,abc=-1$

Now $(a-b)^2+(b-c)^2+(c-a)^2=2(a+b+c)^2-6(ab+bc+ca)=-6$

Could some help me to explain short way to calculate product,Thanks

Answer 1

$a,b$ are the solutions of $x^3+x+1=0$, so we have $$a^3+a=-1=b^3+b$$ and so, $$(a^3-b^3)+(a-b)=0\implies (a-b)(a^2+ab+b^2+1)=0$$

Suppose here that $a=b$. Then, since $a+b+c=0\implies c=-2a$, we have $$abc=-1\implies -2a^3=-1\implies a^3=\frac{1}{2}\implies a=-1-a^3=-\frac 32$$ However, $a=-\frac 32$ is not a solution of $x^3+x+1=0$.

So, we have $a\not=b$, so $$a^2+ab+b^2+1=0\implies (a-b)^2=-3ab-1$$ Similarly, $$(b-c)^2=-3bc-1\qquad\text{and}\qquad (c-a)^2=-3ca-1$$

It follows that $$\begin{align}&(a-b)^2(b-c)^2+(b-c)^2(c-a)^2+(c-a)^2(a-b)^2\\\\&=(-3ab-1)(-3bc-1)+(-3bc-1)(-3ca-1)+(-3ca-1)(-3ab-1)\\\\&=9abc(a+b+c)+6(ab+bc+ca)+3\\\\&=9\cdot (-1)\cdot 0+6\cdot 1+3\\\\&=9\end{align}$$ and $$\begin{align}(a-b)^2(b-c)^2(c-a)^2&=(-3ab-1)(-3bc-1)(-3ca-1)\\\\&=-27(abc)^2-9abc(a+b+c)-3(ab+bc+ca)-1\\\\&=-27\cdot (-1)^2-9\cdot (-1)\cdot 0-3\cdot 1-1\\\\&=-31\end{align}$$

Answer 2

You already know the value of $(a-b)^2+(b-c)^2+(c-a)^2$. Use the fact that\begin{multline}(a-b)^2(b-c)^2(c-a)^2=-4 a b c (a+b+c)^3+(a b+c b+a c)^2 (a+b+c)^2+\\+18 a b c (a b+c b+a c) (a+b+c)-4 (a b+c b+a c)^3-27 a^2 b^2 c^2\end{multline}and that\begin{multline}(a-b)^2(b-c)^2+(b-c)^2(c-a)^2+(c-a)^2(a-b)^2=(a+b+c)^4+\\-6 (a b+c b+a c) (a+b+c)^2+9 (a b+c b+a c)^2\end{multline}to get the other coefficients. This is easy, since you know $a+b+c$, $ab+bc+ca$ and $abc$.

Answer 3

There is the following formula.

Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.

Thus, $$\prod_{cyc}(a-b)^2=27(3u^2v^4-4v^6-4u^3w^3+6uv^2w^3-w^6)=27\left(-\frac{4}{27}-1\right)=-31.$$