Question

If $\alpha ,\beta$ are the root of quadratic equation $a{x}^2+bx+c=0$, then ${(a\alpha +b)}^{-3}+{(a\beta +b)}^{-3}=$

• A. $\frac{b^2-2abc}{a^3{c}^3}$
• B. $\frac{b^3-2abc}{a^2{c}^2}$
• C. $\frac{b^3-3abc}{a^3{c}^3}$
• D. None of these

Answer 1

The correct option is C $\frac{b^3-3abc}{a^3{c}^3}$

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

$\therefore a{\alpha }^2+b\alpha +c=0\Rightarrow a\alpha +b=-\frac{c}{\alpha }$

$\therefore a{\beta }^2+b\beta +c=0\Rightarrow a\beta +b=-\frac{c}{\beta }$

$\therefore {(a\alpha +b)}^{-3}+{(a\beta +b)}^{-3}=(\frac{-c}{\alpha }{)}^{-3}+(\frac{-c}{\beta }{)}^{-3}=-\frac{{\alpha }^3+{\beta }^3}{c^3}$

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

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

${\left(\frac{-b}{a}\right)}^3={\alpha }^3+{\beta }^3+3\left(\frac{c}{a}\right)\left(\frac{-b}{a}\right)$

on solving above expression we get,

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