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Question Number 192186 by Tawa11 last updated on 11/May/23

If α, β and γ are the roots of x^{3} + px + q = 0, find Σ α^{4} .

Answered by BaliramKumar last updated on 10/May/23

{2 p}^{2}

Commented by Tawa11 last updated on 10/May/23

Please workings sir ?

Answered by manxsol last updated on 11/May/23

α + β + γ = 0

α β + α γ + β γ = p

α β γ = - q

α^{4} + p α^{2} + q α = 0

β^{4} + p β^{2} + q β = 0

γ^{4} + p γ^{2} + q γ = 0

α^{4} + β^{4} + γ^{4} + p (α^{2} + β^{2} + γ^{2}) + q (α + β + γ = 0

{(α + β + γ)}^{2} = α^{2} + β^{2} + γ^{2} + 2 (α β + α γ + β γ) = 0

{(0)}^{2} = α^{2} + β^{2} + γ^{2} + 2 (p)

- 2 p = α^{2} + β^{2} + γ^{2}

α^{4} + β^{4} + γ^{4} + p (- 2 p) + q (0) = 0

α^{4} + β^{4} + γ^{4} = 2 p^{2}

Commented by Tawa11 last updated on 11/May/23

God bless you sir .

Answered by BaliramKumar last updated on 11/May/23

α + β + γ = 0

α β + β γ + γ α = p

α β γ = - q

Σ α^{4} = α^{4} + β^{4} + γ^{4} = {(α^{2})}^{2} + {(β^{2})}^{2} + {(γ^{2})}^{2}

\Rightarrow {(α^{2} + β^{2} + γ^{2})}^{2} - 2 (α^{2} β^{2} + β^{2} γ^{2} + γ^{2} α^{2})

\Rightarrow {{(α + β + γ)}^{2} - 2 (α β + β γ + γ α)}^{2}

- 2 {{(α β + β γ + γ α)}^{2} - 2 (α β^{2} γ + α β γ^{2} + α^{2} β γ)}

\Rightarrow {{(α + β + γ)}^{2} - 2 (α β + β γ + γ α)}^{2}

- 2 {{(α β + β γ + γ α)}^{2} - 2 α β γ (α + β + γ)}

\Rightarrow {{(0)}^{2} - 2 (p)}^{2} - 2 {{(p)}^{2} - 2 (- q) (0)}

\Rightarrow {- 2 p}^{2} - 2 {p^{2} - 0}

\Rightarrow {4 p}^{2} - {2 p}^{2}

\Rightarrow \begin{matrix} {2 p}^{2} \end{matrix}

Commented by Tawa11 last updated on 11/May/23

God bless you sir .

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