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

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