Question

If $\alpha ,\beta ,\gamma$ are the roots of the equation $x^3+ax+b=0,$ then $\frac{{\alpha }^3+{\beta }^3+{\gamma }^3}{{\alpha }^2+{\beta }^2+{\gamma }^2}$ is equal to

• A. $\frac{3b}{2a}$
• B. $\frac{-3b}{2a}$
• C. $3b$
• D. $2a$

Answer 1

The correct option is A $\frac{3b}{2a}$

As $\alpha ,\beta ,\gamma$ are roots of $x^3+ax+b=0$
Then $s_1=\alpha +\beta +\gamma =0$
$\Rightarrow {\alpha }^3+{\beta }^3+{\gamma }^3=3\alpha \beta \gamma =-3b$
And ${\alpha }^2+{\beta }^2+{\gamma }^2={(\sum \alpha )}^2-2\sum \alpha \beta =0-2a=-2a$
Therefore $\frac{{\alpha }^3+{\beta }^3+{\gamma }^3}{{\alpha }^2+{\beta }^2+{\gamma }^2}=\frac{-3b}{-2a}=\frac{3b}{2a}$
Hence, option 'A' is correct.