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Question Number 70757 by MJS last updated on 08/Oct/19

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Commented by TawaTawa last updated on 07/Oct/19

$Sir, help me with the question number of a question you solved$ $sometimes .$ $If a + b + c = α$ $a^{2} + b^{2} + c^{2} = β$ $a^{3} + b^{3} + c^{3} = γ$ $Find a^{5} + b^{5} + c^{5}, something like this$
Commented by MJS last updated on 07/Oct/19

$I {can}^{'} t find it now .$ $the idea is : put b = x - y and c = x + y which$ $leads to$ $(1) a + 2 x = α$ $(2) a^{2} + 2 x^{2} + 2 y^{2} = β$ $(3) a^{3} + 2 x^{3} + 6 {x y}^{2} = γ$ $===========$ $(1) \Rightarrow a = α - 2 x$ $\Rightarrow$ $(2) 6 x^{2} - 4 α x + 2 y^{2} + α^{2} = β$ $(3) - 6 x^{3} + 12 α x^{2} + 6 (y^{2} - α^{2}) x + α^{3} = γ$ $====================$ $(2) \Rightarrow y^{2} = - 3 x^{2} + 2 α x + \frac{β - α^{2}}{2}$ $\Rightarrow$ $(3) - 24 x^{3} + 24 α x^{2} - 3 (3 α^{2} - β) x + α^{3} - γ = 0$ $\Rightarrow x^{3} - α x^{2} + \frac{3 α^{2} - β}{8} x - \frac{α^{3} - γ}{24} = 0$ $but$ $a^{4} + b^{4} + c^{4} =$ $= - 32 α (x^{3} - α x^{2} + \frac{3 α^{2} - β}{8} x - \frac{3 α^{4} - 2 α^{2} β + β^{2}}{64 α})$ $and$ $a^{5} + b^{5} + c^{5} =$ $= - 20 (α^{2} + β) (x^{3} + α x^{2} + \frac{3 α^{2} - β}{8} x - \frac{α^{5}}{20 (α^{2} + β)})$ $so we {don}^{'} t have to solve, just compare the$ $constant factors$
Commented by TawaTawa last updated on 07/Oct/19

$God bless you sir, i appreciate$
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