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?
Best Answer
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.