Question

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

• A. $q^3+3pqr+3r^3$
• B. $q^3-3pqr+3r^3$
• C. $q^3+3pqr+3r^2$
• D. $q^3-3pqr+3r^2$

Answer 1

The correct option is C $q^3+3pqr+3r^2$

$\alpha ,\beta$ and $\gamma$ are the roots of the equation $x^3+p{x}^2+qx+r=0$

Sum of the roots taken one at a time $=\sum \alpha =\alpha +\beta +\gamma =-p$

Sum of the roots taken two at a time $=\sum \alpha \beta =\alpha \beta +\beta \gamma +\gamma \alpha =q$

Product of the roots$=-r$

$\sum {\alpha }^3{\beta }^3={\left(\alpha \beta \right)}^3+{\left(\beta \gamma \right)}^3+{\left(\gamma \alpha \right)}^3$

We know that $a^3+b^3+c^3=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)+3abc$

${\left(\alpha \beta \right)}^3+{\left(\beta \gamma \right)}^3+{\left(\gamma \alpha \right)}^3=(\alpha \beta +\beta \gamma +\gamma \alpha )({\left(\alpha \beta \right)}^2+{\left(\beta \gamma \right)}^2+{\left(\gamma \alpha \right)}^2-\alpha {\beta }^2\gamma -\beta {\gamma }^2\alpha -\gamma {\alpha }^2\beta )+3\left(\alpha \beta \right)\left(\beta \gamma \right)\left(\gamma \alpha \right)$

$=\left(q\right)({(\alpha \beta +\beta \gamma +\gamma \alpha )}^2-2\alpha {\beta }^2\gamma -2\beta {\gamma }^2\alpha -2\gamma {\alpha }^2\beta -\alpha {\beta }^2\gamma -\beta {\gamma }^2\alpha -\gamma {\alpha }^2\beta )+3{\left(\alpha \beta \gamma \right)}^2$

$=\left(q\right)({(\alpha \beta +\beta \gamma +\gamma \alpha )}^2-3\alpha {\beta }^2\gamma -3\beta {\gamma }^2\alpha -3\gamma {\alpha }^2\beta )+3r^2$

$=\left(q\right)(q^2-3\alpha \beta \gamma (\alpha +\beta +\gamma ))+3r^2$

$=\left(q\right)(q^2+3rp)+3r^2$

$=q^3+3pqr++3r^2$