Question

If $\alpha +\beta +\gamma =0,$ ${\alpha }^3+{\beta }^3+{\gamma }^3=12$ and ${\alpha }^5+{\beta }^5+{\gamma }^5=40,$ then the value of ${\alpha }^4+{\beta }^4+{\gamma }^4$ is equal to
(correct answer + 1, wrong answer - 0.25)

• A. $8$
• B. $-8$
• C. $4$
• D. $-4$

Answer 1

The correct option is A $8$

As $\alpha +\beta +\gamma =0$. so,cubic equation having $\alpha ,\beta ,\gamma$ as roots is $x^3+px+q=0\cdots \left(1\right)$

$\therefore \alpha \beta +\beta \gamma +\gamma \alpha =p,\alpha \beta \gamma =-q$
$\because \alpha +\beta +\gamma =0$
$\Rightarrow {\alpha }^3+{\beta }^3+{\gamma }^3=3\alpha \beta \gamma$
$\Rightarrow \alpha \beta \gamma =\frac{{\alpha }^3+{\beta }^3+{\gamma }^3}{3}=\frac{12}{3}=4$
$\Rightarrow q=-4$

Also, $\alpha +\beta +\gamma =0$
$\Rightarrow {\alpha }^2+{\beta }^2+{\gamma }^2=-2p$

$\because \alpha ,\beta ,\gamma$ are roots of $x^3+px+q=0$

$\Rightarrow {\alpha }^3+p\alpha +q=0\Rightarrow {\alpha }^5+p{\alpha }^3+q{\alpha }^2=0$
$\Rightarrow \sum {\alpha }^5+p\sum {\alpha }^3+q\sum {\alpha }^2=0$
$\Rightarrow 40+p\left(12\right)+q(-2p)=0\Rightarrow p=-2(\because q=-4)$

$\therefore$ The cubic equation reduces to $x^3-2x-4=0$
$\Rightarrow {\alpha }^3-2\alpha -4=0$
$\Rightarrow {\alpha }^4-2{\alpha }^2-4\alpha =0$

$\sum {\alpha }^4=2\sum {\alpha }^2+4\sum \alpha =2\times (-2p)+4\left(0\right)=-4p$
$=(-4)(-2)=8$

$\therefore {\alpha }^4+{\beta }^4+{\gamma }^4=8$