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

Question

Dr. Abubacker · Accepted Answer

(d)
Here, $\sum_{}^{}{\alpha = 0,\ \ \ \sum_{}^{}{\alpha\beta = b\ \ \ and\ \ \ \alpha\beta\gamma = - c}}$ ...(i)
Now, $\sum_{}^{}{\alpha\ \sum_{}^{}{\alpha\beta = (\alpha + \beta + \gamma).(\alpha\beta + \beta\gamma + \gamma\alpha)}}$
$= \sum_{}^{}{\alpha^{2}\beta + 3\alpha\beta\gamma}$
$\Rightarrow \sum_{}^{}{\alpha^{2}\beta = \sum_{}^{}{\alpha\sum_{}^{}{\alpha\beta - 3\alpha\beta\gamma = 0.\sum_{}^{}{\alpha\beta - 3( - c)}}}}$
[from Eq. (i)]
$= 3c$

Q. If $\alpha,\beta,\gamma$ are the roots of $x^{3} + bx + c = 0$, then $\alpha^{2}\beta + \alpha\beta^{2} + \beta^{2}\gamma + \beta\gamma^{2} + \gamma^{2}\alpha + \gamma\alpha^{2}$ is equal to

Explanation

Reviewed by Dr. Abubacker

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