Browse Questions

# Given $\alpha,\beta,\gamma$ such that $\alpha+\beta+\gamma=2,\alpha^2+\beta^2+\gamma^2=6,\alpha^3+\beta^3+\gamma^3=8$,then the value of $\alpha^4+\beta^4+\gamma^4$ is equal to

$\begin{array}{1 1}(A)\;5&(B)\;12\\(C)\;18&(D)\;36\end{array}$

$(\alpha+\beta+\gamma)^2=\alpha^2+\beta^2+\gamma^2+2(\alpha\beta+\beta\gamma+\alpha\gamma)$
$\Rightarrow 4=6+2(\alpha\beta+\beta\gamma+\alpha\gamma)$
$\Rightarrow \alpha\beta+\beta\gamma+\alpha\gamma=-1$
$(\alpha^3+\beta^3+\gamma^3-3\alpha\beta\gamma=(\alpha+\beta+\gamma)(\alpha^2+\beta^2+\gamma^2-\alpha\beta-\beta\gamma-\alpha\gamma)$
$\Rightarrow 8-3\alpha\beta\gamma=2(6+1)$
$\Rightarrow \alpha\beta\gamma=-2$
$(\alpha^2+\beta^2+\gamma^2)^2=\alpha^4+\beta^4+\gamma^4+2(\alpha^2\beta^2+\beta^2\gamma^2+\alpha^2\gamma^2)$
$\Rightarrow \alpha^4+\beta^3+\gamma^4+2[(\alpha\beta+\beta\gamma+\alpha\gamma)^2-2\alpha\beta\gamma(\alpha+\beta+\gamma)]$
$\Rightarrow 36=\alpha^4+\beta^4+\gamma^4+2[(-1)^2-2(-2(2)]$
$\Rightarrow \alpha^4+\beta^4+\gamma^4=36-18$
$\Rightarrow 18$
Hence (C) is the correct answer.