Search any question & find its solution
Question:
Answered & Verified by Expert
If $\alpha, \beta, \gamma$ are the roots of the equation $x^3-6 x^2+11 x+6=0$, then $\Sigma \alpha^2 \beta+\Sigma \alpha \beta^2$ is equal to :
Options:
Solution:
2013 Upvotes
Verified Answer
The correct answer is:
84
$\because \alpha, \beta, \gamma$ are the roots of the equation
$x^3-6 x^2+11 x+6=0$$
$\therefore \quad \alpha+\beta+\gamma=6$
$\alpha \beta+\beta \gamma+\gamma \alpha=11$
and $\quad \alpha \beta \gamma=-6$
Now $\quad \Sigma \alpha^2 \beta+\Sigma \alpha \beta^2$
$=\alpha^2 \beta+\beta^2 \gamma+\gamma^2 \alpha+\alpha \beta^2+\beta \gamma^2+\gamma \alpha^2$
$=\alpha \beta(\alpha+\beta)+\beta \gamma(\beta+\gamma)+\gamma \alpha(\gamma+\alpha)$
$=\alpha \beta(6-\gamma)+\beta \gamma(6-\alpha)+\gamma \alpha(6-\beta)$
$=6(\alpha \beta+\beta \gamma+\gamma \alpha)-3 \alpha \beta \gamma$
$=6(11)+3(6)$
$=66+18=84$
$x^3-6 x^2+11 x+6=0$$
$\therefore \quad \alpha+\beta+\gamma=6$
$\alpha \beta+\beta \gamma+\gamma \alpha=11$
and $\quad \alpha \beta \gamma=-6$
Now $\quad \Sigma \alpha^2 \beta+\Sigma \alpha \beta^2$
$=\alpha^2 \beta+\beta^2 \gamma+\gamma^2 \alpha+\alpha \beta^2+\beta \gamma^2+\gamma \alpha^2$
$=\alpha \beta(\alpha+\beta)+\beta \gamma(\beta+\gamma)+\gamma \alpha(\gamma+\alpha)$
$=\alpha \beta(6-\gamma)+\beta \gamma(6-\alpha)+\gamma \alpha(6-\beta)$
$=6(\alpha \beta+\beta \gamma+\gamma \alpha)-3 \alpha \beta \gamma$
$=6(11)+3(6)$
$=66+18=84$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.