If $\alpha,\beta,\gamma$ are the roots of $x^3+x+1=0$, then find the equation whose roots are: $(\alpha-\beta)^2,(\beta-\gamma)^2,(\gamma-\alpha)^2$ Question:

If $\alpha,\beta,\gamma$ are the roots of the equation, $x^3+x+1=0$, then find the equation whose roots are: $({\alpha}-{\beta})^2,({\beta}-{\gamma})^2,({\gamma}-{\alpha})^2$

Now, the normal way to solve this question would be to use the theory of equations and find the sum of roots taken one at a time, two at a time and three at a time. Using this approach, we get the answer as $(x+1)^3+3(x+1)^2+27=0$. However, I feel that this is a very lengthy approach to this problem. Is there an easier way of doing it?

The standard method:
$$a+b+c=0; ab+bc+ca=1; abc=-1;\\
a^2+b^2+c^2=-2;a^2b^2+b^2c^2+c^2a^2=1;a^4+b^4+c^4=2;\\
a^3=-a-1.$$
First coefficient:
$$(a-b)^2+(b-c)^2+(c-a)^2=\\2(a^2+b^2+c^2)-2(ab+bc+ca)=-6$$
Second coefficient:
$$(a-b)^2(b-c)^2+(b-c)^2(c-a)^2+(c-a)^2(a-b)^2=\\
a^4+b^4+c^4+3(a^2b^2+b^2c^2+c^2a^2)-\\
2(a^3(\underbrace{b+c}_{-a})+b^3(\underbrace{c+a}_{-b})+c^3(\underbrace{a+b}_{-c}))=\\
3(a^2b^2+b^2c^2+c^2a^2+a^4+b^4+c^4)=9$$
Third coefficient:
$$(a-b)^2(b-c)^2(c-a)^2=\\
\small{(a^2+b^2-2ab)(b^2+c^2-2bc)(c^2+a^2-2ac)=\\
(c^2-4ab)(a^2-4bc)(b^2-4ac)=\\
16 a^4 b c - 4 a^3 b^3 - 4 a^3 c^3 - 63 a^2 b^2 c^2 + 16 a b^4 c + 16 a b c^4 - 4 b^3 c^3=\\
-16(a^3+b^3+c^3)-4(a^3b^3+b^3c^3+c^3a^3)-63=\\
-16(-a-b-c-3)-4((-a-1)(-b-1)+\\
(-b-1)(-c-1)+(-c-1)(-a-1))-63=}\\
48-4(1+3)-63=-31.$$
Hence, the equation is: $x^3+6x^2+9x+31=0$.
