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Question Number 64066 by mathmax by abdo last updated on 12/Jul/19

let α ,β and λ the roots of x^3 +2x−1 =0 find the value of  A =α^2  +β^2  +λ^2  and  B =α^3  +β^3  +λ^3  .

$${let}\:\alpha\:,\beta\:{and}\:\lambda\:{the}\:{roots}\:{of}\:{x}^{\mathrm{3}} +\mathrm{2}{x}−\mathrm{1}\:=\mathrm{0}\:{find}\:{the}\:{value}\:{of} \\ $$$${A}\:=\alpha^{\mathrm{2}} \:+\beta^{\mathrm{2}} \:+\lambda^{\mathrm{2}} \:{and}\:\:{B}\:=\alpha^{\mathrm{3}} \:+\beta^{\mathrm{3}} \:+\lambda^{\mathrm{3}} \:. \\ $$

Answered by MJS last updated on 12/Jul/19

the solutions are  α=u+v  β=(−(1/2)−((√3)/2)i)u+(−(1/2)+((√3)/2)i)v  λ=(−(1/2)+((√3)/2)i)u+(−(1/2)−((√3)/2)i)v  A=6uv  B=3(u^3 +v^3 )  u=((−(q/2)+((√(3(4p^3 +27q^2 ))/(18))))^(1/3)   v=((−(q/2)−((√(3(4p^3 +27q^2 ))/(18))))^(1/3)   A=−2p  B=−3q  p=2  q=−1  A=−4  B=3

$$\mathrm{the}\:\mathrm{solutions}\:\mathrm{are} \\ $$$$\alpha={u}+{v} \\ $$$$\beta=\left(−\frac{\mathrm{1}}{\mathrm{2}}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{i}\right){u}+\left(−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{i}\right){v} \\ $$$$\lambda=\left(−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{i}\right){u}+\left(−\frac{\mathrm{1}}{\mathrm{2}}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{i}\right){v} \\ $$$${A}=\mathrm{6}{uv} \\ $$$${B}=\mathrm{3}\left({u}^{\mathrm{3}} +{v}^{\mathrm{3}} \right) \\ $$$${u}=\sqrt[{\mathrm{3}}]{−\frac{{q}}{\mathrm{2}}+\frac{\sqrt{\mathrm{3}\left(\mathrm{4}{p}^{\mathrm{3}} +\mathrm{27}{q}^{\mathrm{2}} \right.}}{\mathrm{18}}} \\ $$$${v}=\sqrt[{\mathrm{3}}]{−\frac{{q}}{\mathrm{2}}−\frac{\sqrt{\mathrm{3}\left(\mathrm{4}{p}^{\mathrm{3}} +\mathrm{27}{q}^{\mathrm{2}} \right.}}{\mathrm{18}}} \\ $$$${A}=−\mathrm{2}{p} \\ $$$${B}=−\mathrm{3}{q} \\ $$$${p}=\mathrm{2} \\ $$$${q}=−\mathrm{1} \\ $$$${A}=−\mathrm{4} \\ $$$${B}=\mathrm{3} \\ $$

Commented by mathmax by abdo last updated on 12/Jul/19

thank you sir mjs

$${thank}\:{you}\:{sir}\:{mjs} \\ $$

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