Question

If $\alpha ,\beta ,\gamma$ are such that $\alpha +\beta +\gamma =2$ ${\alpha }^2+{\beta }^2+{\gamma }^2=6,{\alpha }^3+{\beta }^3+{\gamma }^3=8$, then ${\alpha }^4+{\beta }^4+{\gamma }^4$ is equal to

• A. 7
• B. 12
• C. 18
• D. 36

Answer 1

Answer: (C)

Given, $\alpha +\beta +\gamma =2,$
${\alpha }^2+{\beta }^2+{\gamma }^2=6$,
${\alpha }^3+{\beta }^3+{\gamma }^3=8$
Now, $(\alpha +\beta +\gamma {)}^2=2^2$
$\Rightarrow {\alpha }^2+{\beta }^2+{\gamma }^2+2(\alpha \beta +\beta \gamma +\gamma \alpha )=4$
$\Rightarrow 2(\alpha \beta +\beta \gamma +\gamma \alpha )=4-6=-2$
Also, ${\alpha }^3+{\beta }^3+{\gamma }^3-3\alpha \beta \gamma$
$=(\alpha +\beta +\gamma )\left({\alpha }^2+{\beta }^2+{\gamma }^2-\alpha \beta -\beta \gamma -\gamma \alpha \right)$
$\Rightarrow 8-3\alpha \beta \gamma =2[6-(-1)]$
$\Rightarrow 8-3\alpha \beta \gamma =14$
$\Rightarrow \alpha \beta \gamma =8-14$
$\Rightarrow \alpha \beta \gamma =-2$
Now
${\alpha }^4+{\beta }^4+{\gamma }^4$
$={\left({\alpha }^2+{\beta }^2+{\gamma }^2\right)}^2-2\Sigma {\alpha }^2{\beta }^2$
$={\left({\alpha }^2+{\beta }^2+{\gamma }^2\right)}^2-2\left[(\Sigma \beta \gamma {)}^2-2\alpha \beta \gamma \Sigma \alpha \right]$
$=(6{)}^2-2\left[(-1{)}^2-2(-2)2\right]$
$=36-2[9]$
$=36-18$
$=18$