Question

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

• A. $\frac{(b^3–3abc)}{a^3{c}^3}$
• B. $a^3-2abc$
• C. $b^3–3abc$
• D. $\frac{(c^3–3abc)}{b^3{c}^3}$

Answer 1

The correct option is A

$\frac{(b^3–3abc)}{a^3{c}^3}$

α+β= $\frac{-b}{a}$ , αβ=$\frac{c}{a}$

α ≠ 0 is root of the equation $a{x}^2+bx+c$ =0 then $a{\alpha }^2+b\alpha +c$=0 => aα+b=$\frac{c}{a}$

similarly αβ+b = $\frac{-c}{\beta }$

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