Explanation

(b)
Here, $\sum_{}^{}{\alpha = 0,\ \ \ \ \ \sum_{}^{}{\alpha\beta = - 7,\ \ \ \ \ \alpha\beta\gamma = - 7}}$
$\therefore\ \ \frac{1}{\alpha^{4}} + \frac{1}{\beta^{4}} + \frac{1}{\gamma^{4}} = \frac{\alpha^{4}\beta^{4} + \beta^{4}\gamma^{4} + \gamma^{4}\beta^{4}}{\alpha^{4}\beta^{4}\gamma^{4}}$
$= \frac{\sum_{}^{}{\alpha^{4}\beta^{4}}}{\alpha^{4}\beta^{4}\gamma^{4}}$ ...(i)
Now, $\sum_{}^{}{\alpha\beta\sum_{}^{}{\alpha\beta\sum_{}^{}{\alpha\beta\sum_{}^{}{\alpha\beta = \left( \sum_{}^{}{\alpha\beta} \right)^{2}\left( \sum_{}^{}{\alpha\beta} \right)^{2}}}}}$
$\Rightarrow ( - 7)^{2} = \left\lbrack \alpha^{2}\beta^{2} + \beta^{2}\gamma^{2} + \gamma^{2}\alpha^{2} + 2\alpha\beta\gamma(\alpha + \beta + \gamma) \right\rbrack\lbrack\alpha^{2}\beta^{2} + \beta^{2}\gamma^{2} + \gamma^{2}\alpha^{2} + 2\alpha\beta\gamma(\alpha + \beta + \gamma)\rbrack$
$= \left( \alpha^{4}\beta^{4} + \beta^{4}\gamma^{4} + \gamma^{4}\alpha^{4} \right)\left( \alpha^{4}\beta^{4} + \beta^{4}\gamma^{4} + \gamma^{4}\alpha^{4} \right)\ \ \left\lbrack \sum_{}^{}{\alpha = \alpha + \beta + \gamma = 0} \right\rbrack$
$= \alpha^{4}\beta^{4} + \beta^{4}\gamma^{4} + \gamma^{4}\alpha^{4} + 2\alpha^{4}\beta^{2}\gamma^{2} + 2\alpha^{2}\beta^{4}\gamma^{2} + 2\alpha^{2}\beta^{2}\gamma^{4}$
$= \sum_{}^{}{\alpha^{4}\beta^{4} + 2\alpha^{2}\beta^{2}\gamma^{2}\lbrack\left( \sum_{}^{}\alpha \right)^{2} - 2\sum_{}^{}{\alpha\beta}\rbrack}$
$= \sum_{}^{}{\alpha^{4}\beta^{4} + 2\alpha^{2}\beta^{2}\gamma^{2}\left\lbrack 0 - 2 \times ( - 7) \right\rbrack}$
$= \sum_{}^{}{\alpha^{4}\beta^{4} + 2( - 7)^{2}(2 \times 7)}$
$\Rightarrow \sum_{}^{}{\alpha^{4}\beta^{4} = ( - 7)^{4} + 4( - 7)^{3}}$
$\Rightarrow \ \ \sum_{}^{}{\alpha^{4}\beta^{4} = ( - 7)^{3}( - 7 + 4) = - 3( - 7)^{3}}$
On putting this value in Eq. (i), we get
$\frac{1}{\alpha^{2}} + \frac{1}{\beta^{4}} + \frac{1}{\gamma^{4}} = \frac{- 3( - 7)^{3}}{( - 7)^{4}} = \frac{- 3}{- 7} = \frac{3}{7}$