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

Let $a,b,c$ be the roots of $x^3+x+1=0$ so we have $a+b+c=0, ab+bc+ca=1,abc=-1$, so $a^2+b^2+c^2=-2$ and $c^3=-c-1$

We would explore a transformation from $x$ to $y$ to get the required cubic equation of $y$. Let $$y=(a-b)^2=a^2+b^2-2ab=y=-2-c^2+2/c \implies c=\frac{3}{1+y}$$ Replacing $c$ by $x$ we get the required transformation $x=\frac{3}{1+y}$, putting it in the given $x$ equation, we get: $$\frac{27}{(1+y)^3}+\frac{3}{(1+y)}+1=0 \implies y^3+6y^2+9y+31=0,$$ which is the required cubic equation.

Final constant term $1+3+27=31$ can be obtained at once (or checked at once) by considering that it is the opposite of the product of roots

$$(({\alpha}-{\beta})({\beta}-{\gamma})({\gamma}-{\alpha}))^2$$

which is the classical discriminant $-(4p^3+27q^2)$ of a reduced 3rd degree equation $X^3+pX+q=0$ with $p=q=1$. (https://en.wikipedia.org/wiki/Discriminant#Degree_3)

Hint:

Let $y=(a-b)^2=(a+b)^2-4ab=(-c)^2-\dfrac4{-c}$ as $abc=-1, a+b=-c$

$$\iff c^3-cy+4=0\ \ \ \ (1) $$

Again we have $$c^3+c+1=0\ \ \ \ (0)$$

Solve the two simultaneous equations for $c,c^3$ and use $c^3=(c)^3$ to eliminate $c$