Question

Form the equation whose roots are the squares of the differences of the roots of the cubic
$x^3+qx+r=0$.

Answer 1

Let $a,b,c$ be the roots of the cubic $x^3+qx+r=0$; then the roots of the required equation are $(b-c{)}^2,(c-a{)}^2,(a-b{)}^2$.

Also, $a+b+c=0,ab+bc+ac=q$ and $abc=-r$
Now $(b-c{)}^2=b^2+c^2-2bc=a^2+b^2+c^2-a^2-\frac{2abc}{a}$
$=(a+b+c{)}^2-2(bc+ca+ab)-a^2-\frac{2abc}{a}$
$=-2q-a^2+\frac{2r}{a}$;
Also when $x=a$ in the given equation, $y=(b-c{)}^2$ in the transformed equation;
$\therefore y=-2q-x^2+\frac{2r}{x}$.
Thus we have to eliminate $x$ between the equations
$x^3+qx+r=0$,
and $x^3+(2q+y)x-2r=0$.
By subtraction $(q+y)x=3r$; or $x=\frac{3r}{q+y}$.
Substituting and reducing, we obtain
$y^3+6q{y}^2+9q^{2}y+27r^2+4q^3=0$.