Question

If $\alpha ,\beta$ and $\gamma$ are the roots of the equation $x^3-6x^2+11x+6=0$, then $\sum {\alpha }^2\beta +\sum \alpha {\beta }^2$ is equal to

• A. $80$
• B. $168$
• C. $90$
• D. $-84$

Answer 1

The correct option is B

$168$

Explanation for the correct option:

Find the value of $\sum {\alpha }^2\beta +\sum \alpha {\beta }^2$:

Given, $\alpha ,\beta$ and $\gamma$ are the roots of the equation $x^3-6x^2+11x+6=0$,

$\Rightarrow \sum ɑ=6,\sum \alpha \beta =11$ and $ɑ\beta ɣ=–6$

$\begin{array}{rcl}\therefore \sum {\alpha }^2\beta +\sum \alpha {\beta }^2& =& 2({\alpha }^2\beta +{\beta }^2ɣ+{ɣ}^2\alpha +\beta {ɣ}^2+ɣ{\alpha }^2)\\ & =& 2[\alpha \beta (\alpha +\beta )+\beta ɣ(\beta +ɣ)+ɣ\alpha (ɣ+\alpha \left)\right]\\ & =& 2[\alpha \beta (6–ɣ)+\beta ɣ(6–\alpha )+ɣ\alpha (6–\beta \left)\right]\\ & =& 2[6(\alpha \beta +\beta ɣ+ɣ\alpha )–ɑ\beta ɣ]\\ & =& 2[6(11)–3(-6\left)\right]\\ & =& 2(66+18)\\ & =& \mathbf{168}\end{array}$

Hence, Option ‘B’ is Correct.