If $α_{1}, α_{2}, α_{3}, α_{4}, α_{5}$ are roots of equation $x^{5} - 5 x^{4} - 1 = 0$ , then

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$\sum_{r = 1}^{r = 5} \frac{1}{α_{r}^{4}} = - \frac{1}{20}$

$\prod_{r = 1}^{r = 5} {(\frac{1}{α_{r}^{4}} + 5)}^{5} = 1$

$\prod_{r = 1}^{r = 5} {(\frac{1}{α_{r}^{4}} + 5)}^{3} = \frac{1}{5}$

$\sum_{r = 1}^{r = 5} \frac{1}{α_{r}^{4}} = - 20$

answer is B, C.

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Detailed Solution

$x^{4} (x - 5) = 1$
$\Rightarrow x - 5 = \frac{1}{x^{4}}$
$∴ \sum \frac{1}{α_{r}^{4}} = \sum α_{r} - 25 = 5 - 25 = - 20$
Similarly, $(\frac{1}{α_{r}^{4}} + 5) = α_{r}$

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