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If $α + β + γ = 6, α^{2} + β^{2} + γ^{2} = 14$ and $α^{3} + β^{3} + γ^{3} = 36$ then $α^{4} + β^{4} + γ^{4}$ is equal to

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Detailed Solution

$S_{1} = 6, S_{2} = 14, S_{3} = 36$
$\Rightarrow P_{1} = - S_{1} = - 6$ , $P_{2} = \frac{S_{1}^{2} - S_{2}}{2} = 11$ and $P_{3} = \frac{S_{1} (S_{2} - P_{2}) - S_{3}}{3} = - 6$

Newtons formula is $S_{4} + P_{1} S_{3} + P_{2} S_{2} + P_{3} S_{1} = 0$

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