Question

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

• 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 $a{x}^2+bx+c=0$

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

$\Rightarrow$ ${\alpha }^3{\beta }^3=\frac{c^3}{a^3}$ ----- ( 2 )

$\Rightarrow$ $\alpha +\beta =\frac{-b}{a}$ ------ ( 3 )

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

$\Rightarrow$ ${\left(\frac{-b}{a}\right)}^3={\alpha }^3+{\beta }^3+3\left(\frac{c}{a}\right)\left(\frac{-b}{a}\right)$ [ By using ( 1 ) and ( 3 ) ]

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

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

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

$\therefore$ ${\alpha }^3+{\beta }^3=\frac{3abc-b^3}{a^3}$ ----- ( 4 )

Now,

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

$=\frac{\frac{3abc-b^3}{a^3}}{\frac{c^3}{a^3}}$ [ By using ( 2 ) and ( 4 ) ]

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

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

$\therefore$ $\frac{1}{{\alpha }^3}+\frac{1}{{\beta }^3}==\frac{3abc-b^3}{c^3}$