Question

Find the quadratic equation, whose sum of roots is $3$ and sum of their cubes is $63$.

Answer 1

Let $\alpha ,\beta$ be roots of equation

$\alpha +\beta =3$

${\alpha }^3+{\beta }^3=63$

${\alpha }^3+{\beta }^3=(\alpha +\beta {)}^3-3\alpha \beta (\alpha +\beta )$

$6^3=(3{)}^3-3\alpha \beta (3)$

$63=27-9\alpha \beta$

$9\alpha \beta =-36$

$\alpha \beta =-4$

$\therefore$ equation

$x^2-3x+(-4)=0$