Explanation

(c)
Given, $\alpha + \beta + \gamma = 2,\ \alpha^{2} + \beta^{2} + \gamma^{2} = 6$,
$\alpha^{3} + \beta^{3} + \gamma^{2} = 8$
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, $\alpha^{3} + \beta^{3} + \gamma^{3} - 3\alpha\beta\gamma$
$= (\alpha + \beta + \gamma)(\alpha^{2} + \beta^{2} + \gamma^{2} - \alpha\beta - \beta\gamma - \gamma\alpha)$
$\Rightarrow \ \ \ 8 - 3\alpha\beta\gamma = 2\lbrack 6 - ( - 1)\rbrack$
$\Rightarrow 8 - 3\alpha\beta\gamma = 14$
$\Rightarrow 3\alpha\beta\gamma = 8 - 14$
$\Rightarrow \ \alpha\beta\gamma = - 2$
Now, $\alpha^{4} + \beta^{4} + \gamma^{4} = \left( \alpha^{2} + \beta^{2} + \gamma^{2} \right)^{2} - 2\sum_{}^{}{\alpha^{2}\beta^{2}}$
$= \left( \alpha^{2} + \beta^{2} + \gamma^{2} \right)^{2} - 2\left\lbrack \left( \sum_{}^{}{\beta\gamma} \right)^{2} - 2\alpha\beta\gamma\sum_{}^{}\alpha \right\rbrack$
$= (6)^{2} - 2\lbrack( - 1)^{2} - 2( - 2)2\rbrack$
$= 36 - 2\lbrack 9\rbrack$
$= 36 - 18 = 18$