Question

lf $\alpha ,\beta ,\gamma$ are the roots of $x^3-px+q=0$, then ${\alpha }^6+{\beta }^6+r^6$, is

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

Answer 1

The correct option is C $2p^3-3q^3$

We need to find value of ${\alpha }^6+{\beta }^6+{\gamma }^6=3\left({\alpha }^3{\beta }^3{\gamma }^3\right)+({\alpha }^2+{\beta }^2+{\gamma }^2)[{({\alpha }^2+{\beta }^2+{\gamma }^2)}^2-3({\alpha }^2{\beta }^2+{\beta }^2{\gamma }^2+{\gamma }^2{\alpha }^2\left)\right]$

$\beta \gamma \alpha =-q\alpha +\beta +\gamma =0$

$\alpha \beta +\beta \gamma +\gamma \alpha =-p$

${\alpha }^2+{\beta }^2+{\gamma }^2=2p$

${\alpha }^2{\beta }^2+{\beta }^2{\gamma }^2+{\gamma }^2{\alpha }^2=p^2$

${\alpha }^6+{\beta }^6+{\gamma }^6=-3q^3+2p\left[\right(4p^2-3p^2\left)\right]$

${\alpha }^6+{\beta }^6+{\gamma }^6=2p^3-3q^3$