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

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 $abc$.
