Question

If $\alpha$and $\beta$are roots of$4x^2+3x+7=0$, then the value of$\left(\frac{1}{{\alpha }^3}\right)+\left(\frac{1}{{\beta }^3}\right)=$

• A. $\raisebox{1ex}{$-27$}\!\left/ \!\raisebox{-1ex}{$64$}\right.$
• B. $\raisebox{1ex}{$225$}\!\left/ \!\raisebox{-1ex}{$343$}\right.$
• C. $\raisebox{1ex}{$63$}\!\left/ \!\raisebox{-1ex}{$16$}\right.$
• D. $\raisebox{1ex}{$63$}\!\left/ \!\raisebox{-1ex}{$64$}\right.$

Answer 1

The correct option is B

$\raisebox{1ex}{$225$}\!\left/ \!\raisebox{-1ex}{$343$}\right.$

Explanation for correct option:

We have given that$\alpha and\beta$are the roots of the equation$4x^2+3x+7=0$

and need to find the value off$\left(\frac{1}{{\alpha }^3}\right)+\left(\frac{1}{{\beta }^3}\right)=$

$$\begin{gathered} \therefore \alpha +\beta =\frac{-b}{a}=\frac{-3}{4} \\ \alpha \beta =\frac{c}{a}=\frac{7}{4} \\ so, \\ \frac{1}{{\alpha }^3}+\frac{1}{{\beta }^3}=\frac{{\beta }^3+{\alpha }^3}{{\alpha }^3{\beta }^3}=\frac{[{(\beta +\alpha )}^3-3\alpha \beta (\beta +\alpha \left)\right]}{{\left(\alpha \beta \right)}^3} \\ \Rightarrow \frac{\left[\right({\displaystyle \raisebox{1ex}{$-3$}\!\left/ \!\raisebox{-1ex}{$4{)}^3-3\times \left({\displaystyle \frac{7}{4}}\right)(-{\displaystyle \frac{3}{4}})$}\right.}]}{{\left({\displaystyle \frac{7}{4}}\right)}^3}=\frac{225}{343} \end{gathered}$$

Hence, option$\left(B\right)$is correct answer