Question

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

• A. $\frac{3abc-b^3}{a^3}$
• B. $\frac{3ab-b^3}{a^{2}c}$
• C. $\frac{3abc-b^3}{c^3}$
• D. $\frac{b^2-2ac}{ac}$

Answer 1

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

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

$\Rightarrow$ $\alpha +\beta =\frac{-b}{a}$ ------- ( 1 )

$\Rightarrow$ $\alpha \beta =\frac{c}{a}$ ----- ( 2 )

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

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

$\frac{-b^3}{a^3}={\alpha }^3+{\beta }^3+3\times \frac{c}{a}\times \frac{-b}{a}$ [ Using ( 1 ) and ( 2 ) ]

$\Rightarrow$ $\frac{-b^3}{a^3}={\alpha }^3+{\beta }^3-\frac{3bc}{a^2}$

$\therefore$ ${\alpha }^3+{\beta }^3=\frac{3bc}{a^2}-\frac{b^3}{a^3}$

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

Now,

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

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

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

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