Question

If $\alpha ,\beta ,\gamma$ are the roots of $x^3+p{x}^2+qx+r=0$, then $\sum {\alpha }^2(\beta +\gamma )$ is

• A. $3r+pq$
• B. $3r-pq$
• C. $pq-3r$
• D. $pq+r$

Answer 1

The correct option is B $3r-pq$

As $\alpha ,\beta ,\gamma$ are the roots of $x^3+p{x}^2+qx+r=0$
It gives
$\alpha +\beta +\gamma =-p$

$\alpha \beta +\beta \gamma +\gamma \alpha =q$

$\alpha \beta \gamma =-r$

Now,
$\sum {\alpha }^2(\beta +\gamma )=({\alpha }^2\beta +{\alpha }^2\gamma )+({\beta }^2\gamma +{\beta }^2\alpha )+({\gamma }^2\alpha +{\gamma }^2\beta )$

$=(\alpha +\beta +\gamma )(\alpha \beta +\beta \gamma +\gamma \alpha )-3\alpha \beta \gamma$

$=-pq+3r$

$=3r-pq$

Hence, option B.