Question

If $\alpha$,$\beta$ are the roots of $3x^2+7x+5=0$, then the value of ${\left(\frac{1}{3\alpha +7}\right)}^3+{\left(\frac{1}{3\beta +7}\right)}^3$ is

• A. $\frac{-28}{3375}$
• B. $\frac{-672}{3375}$
• C. $\frac{672}{3375}$
• D. $\frac{28}{3375}$
  

Answer 1

The correct option is D $\frac{28}{3375}$


$\alpha$,$\beta$ are the roots of $3x^2+7x+5=0$
$\Rightarrow 3{\alpha }^2+7\alpha +5=0$ and $3{\beta }^2+7\beta +5=0$
$\Rightarrow \frac{1}{3\alpha +7}=\frac{-\alpha }{5}$ and $\frac{1}{3\beta +7}=\frac{-\beta }{5}$
$\therefore {\left(\frac{1}{3\alpha +7}\right)}^3+{\left(\frac{1}{3\beta +7}\right)}^3={\left(\frac{-\alpha }{5}\right)}^3+{\left(\frac{-\beta }{5}\right)}^3$
$=\frac{-1}{125}({\alpha }^3+{\beta }^3)$
$=\frac{-1}{125}(\alpha +\beta )({\alpha }^2-\alpha \beta +{\beta }^2)$
$=\frac{-1}{125}(\alpha +\beta )\left[\right(\alpha +\beta {)}^2-3\alpha \beta ]$
$=\frac{-1}{125}\left(\frac{-7}{3}\right)[\frac{49}{9}-3\times \frac{5}{3}]$
$=\frac{28}{3375}$