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Question Number 63642 by bshahid010@gmail.com last updated on 06/Jul/19

Commented by kaivan.ahmadi last updated on 06/Jul/19

$$(x+2)(x^2-2x+4)=0\Rightarrow$$$$x=-2\Rightarrow \gamma =-2\Rightarrow {\gamma }^2=4$$$$\alpha ,\beta aretherootsof{x}^2-2x+4=0$$$$theequationis$$$$(x-{\alpha }^2)(x-{\beta }^2)(x-4)=0$$$$(x^2-({\alpha }^2+{\beta }^2)x+{\alpha }^2{\beta }^2)(x-4)=0$$$$but$$$$\alpha +\beta =2,\alpha \beta =4$$$${\alpha }^2+{\beta }^2={(\alpha +\beta )}^2-2\alpha \beta =4-8=-4$$$${\alpha }^2{\beta }^2={\left(\alpha \beta \right)}^2=16$$$$sotheequationis$$$$(x^2+4x+16)(x-4)=0\Rightarrow$$$$x^3-4x^2+4x^2-16x+16x-64=0\Rightarrow$$$$x^3-64=0$$$$$$

Answered by MJS last updated on 06/Jul/19

$$x^3-8^2=0$$

Answered by MJS last updated on 06/Jul/19

$$x^3=-8$$$$x_1=-2\Rightarrow x_1^2=4$$$$x_2={2\mathrm{e}}^{\mathrm{i}\frac{\pi }{3}}\Rightarrow x_2^2={4\mathrm{e}}^{\mathrm{i}\frac{2\pi }{3}}$$$$x_3={2\mathrm{e}}^{-\mathrm{i}\frac{\pi }{3}}\Rightarrow x_3^2={4\mathrm{e}}^{-\mathrm{i}\frac{2\pi }{3}}$$$$\mathrm{we}\mathrm{know}\mathrm{that}{x}_1^3=64\Rightarrow \mathrm{the}\mathrm{equation}\mathrm{is}$$$$x^3=64$$$$\mathrm{to}\mathrm{be}\mathrm{sure}:$$$$(x-4)(x-{4\mathrm{e}}^{\mathrm{i}\frac{2\pi }{3}})(x-{4\mathrm{e}}^{-\mathrm{i}\frac{2\pi }{3}})=$$$$=(x-4)(x^2-4x({\mathrm{e}}^{\mathrm{i}\frac{2\pi }{3}}+{\mathrm{e}}^{-\mathrm{i}\frac{2\pi }{3}})+16)=$$$$=(x-4)(x^2-4x(-1)+16)=$$$$=(x-4)(x^2+4x+16)=x^3-64$$

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