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
$\alpha^{4}+\beta^{4}+\gamma^{4}$
$=\left(\alpha^{2}+\beta^{2}+\gamma^{2}\right)^{2}-2 \Sigma \alpha^{2} \beta^{2}$
$=\left(\alpha^{2}+\beta^{2}+\gamma^{2}\right)^{2}-2\left[(\Sigma \beta \gamma)^{2}-2 \alpha \beta \gamma \Sigma \alpha\right]$
$=(6)^{2}-2\left[(-1)^{2}-2(-2) 2\right]$
$=36-2[9]$
$=36-18$
$=18$