Question

Find the sum of the squares and of the cubes of the roots of the equation $x^3-p{x}^2+qx-r=0$.

Answer 1

Denote the roots by $a,b,c$; then
$a+b+c=p,bc+ca+ab=q$.
Now $a^2+b^2+c^2=(a+b+c{)}^2-2(bc+ca+ab)$
$=p^2-2q$.
Again, substitute $a,b,c$ for $x$ in the given equation and add; thus
$a^3+b^3+c^3-p(a^2+b^2+c^2)+q(a+b+c)-3r=0$;
$\therefore a^3+b^3+c^3=p(p^2-2q)-pq+3r$
$=p^3-3pq+3r$.