Question

If $\alpha ,\beta$ are roots of ${ax}^2+bx+c=0$ then $\frac{1}{{\alpha }^3}+\frac{1}{{\beta }^3}=$

Answer 1

Given,

$\alpha ,\beta$ are the roots of $a{x}^2+bx+c=0$

$\alpha +\beta =-\frac{b}{a}$

$\alpha \beta =\frac{c}{a}$

$\frac{1}{{\alpha }^3}+\frac{1}{{\beta }^3}$

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

$=\frac{(\alpha +\beta )({\alpha }^2-\alpha \beta +{\beta }^2)}{(\alpha \beta {)}^3}$

$=\frac{(\alpha +\beta )\left[\right(\alpha +\beta {)}^2-3\alpha \beta ]}{(\alpha \beta {)}^3}$

$=\frac{(-\frac{b}{a})[{(-\frac{b}{a})}^2-3\left(\frac{c}{a}\right)]}{{\left(\frac{c}{a}\right)}^3}$

$=\frac{\frac{-b^3+3abc}{a^3}}{\frac{c^3}{a^3}}$

$=\frac{-b^3+3abc}{c^3}$