Question

If $\alpha ,\beta ,\gamma$ are thhe roots of the equation $x^3+p{x}^2+qx+r=0$ then $\sum {\alpha }^3{\beta }^3$

Answer 1

Put $y=x^3$ in the equation $x^3+r=-p{x}^2-qx$

$\Rightarrow y+r=-x(px+q)$

$\Rightarrow {(y+r)}^3=-x^3{(px+q)}^3$

$\Rightarrow {(y+r)}^3=-y(p^3{x}^3+q^3+3p^{2}q{x}^2+3p{q}^{2}x)$

$\Rightarrow {(y+r)}^3=-y(p^{3}y+q^3+3pq(p{x}^2+qx))$

$\Rightarrow {(y+r)}^3=-y(p^{3}y+q^3+3pq(-x^3-r))$

$\Rightarrow {(y+r)}^3=-y(p^{3}y+q^3+3pq(-y-r))$

$\Rightarrow y^3+(3r+p^3-3pq)y^2+(3r^2+q^3-3pqr)y+r^3=0$

Roots of this equation are ${\alpha }^3,{\beta }^3,{\gamma }^3$

$\therefore \sum {\alpha }^3{\beta }^3=3r^2+q^3-3pqr$