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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$α^{4} + β^{4} + γ^{4} = 98$

$α^{4} + β^{4} + γ^{4} = 89$

$α^{4} + β^{4} + γ^{4} = 79$

$α^{4} + β^{4} + γ^{4} = 97$

answer is A.

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Detailed Solution

$\begin{array}{l} α + β + γ = 6 \\ α^{2} + β^{2} + γ^{2} = 14 \\ α^{3} + β^{3} + γ^{3} = 36 \\ α β + β γ + γ α = \frac{{(α + β + γ)}^{2} - (α^{2} + β^{2} + γ^{2})}{2} \end{array}$

$\begin{array}{l} \sum α β = \frac{{(6)}^{2} - 14}{2} \\ \sum α β = \frac{36 - 14}{2} = \frac{22}{2} \\ \sum α β = 11 \\ \sum α^{3} - 3 α β γ = (\sum α) (\sum α^{2} - \sum α β) \end{array}$

$\begin{array}{l} α β γ = \frac{\sum α^{3} - (\sum α) (\sum α^{2} - \sum α β)}{3} \\ α β γ = \frac{36 - (6) (14 - 11)}{3} \\ α β γ = 12 - 2 (3) \\ α β γ = 6 \end{array}$

Now $α^{3} - (\sum α) x^{2} + (\sum α β) x - α β γ = 0$

$\begin{array}{l} x^{3} - 6 x^{2} + 11 x - 6 = 0 \\ ∴ a_{0} s_{4} + a_{1} s_{3} + a_{2} s_{2} + a_{3} s_{1} + 4 a_{4} = 0 \\ (1) s_{4} + (- 6) (36) + (11) (14) + (- 6) (6) + 4 (0) = 0 \\ s_{4} - 216 + 154 - 36 = 0 \end{array}$