Find the sum of the fourth powers of the roots of $x^{3} + q x + r = 0$ .

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Solution

$x^{3} + q x + r = 0 \sum a = a + b + c = 0 \sum a b = a b + b c + a c = q a b c = - r a^{4} + b^{4} + c^{4} + d^{4} = \sum a^{4} = {(\sum a^{2})}^{2} - 2 \sum a^{2} b^{2} = {(\sum a - 2 \sum a b)}^{2} - 2 ({(\sum a b)}^{2} - 2 a b c \sum a) = {(0 - 2 q)}^{2} - 2 {(q - 0)}^{2} = 4 q^{2} - 2 q^{2} = 2 q^{2}$

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