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Question Number 50925 by kaivan.ahmadi last updated on 22/Dec/18

$$\mathrm{factor}{\mathrm{x}}^3+{\mathrm{x}}^2+\mathrm{x}-\frac{1}{3}\mathrm{and}$$$${\mathrm{x}}^3+{\mathrm{x}}^2-\mathrm{x}+\frac{1}{3}\mathrm{and}$$$${\mathrm{x}}^3+{\mathrm{x}}^2-\mathrm{x}-\frac{1}{3}$$

Commented by Abdo msup. last updated on 23/Dec/18

$$letp\left(x\right)=x^3+x^2+x-\frac{1}{3}letfindtherootschangementx=t-\frac{1}{3}$$$$givep\left(x\right)={(t-\frac{1}{3})}^3+{(t-\frac{1}{3})}^2+t-\frac{2}{3}$$$$=t^3-3t^2.\frac{1}{3}+3t\frac{1}{9}-\frac{1}{27}+t^2-\frac{2}{3}t+\frac{1}{9}+t-\frac{2}{3}$$$$=t^3-t^2+\frac{t}{3}+t^2+\frac{1}{3}t+\frac{-1+3-18}{27}$$$$=t^3+\frac{2}{3}t-\frac{16}{27}=g\left(t\right)letsolveg\left(t\right)=0witht=u+v$$$$weget{(u+v)}^3+\frac{2}{3}(u+v)-\frac{16}{27}=0\Rightarrow$$$$u^3+v^3+3uv(u+v)+\frac{2}{3}(u+v)-\frac{16}{27}=0\Rightarrow$$$$u^3+v^3-\frac{16}{27}=0and3uv+\frac{2}{3}=0\Rightarrow$$$$u^3+v^3=\frac{16}{27}anduv=-\frac{2}{9}\Rightarrow$$$$u^3+v^3=\frac{16}{27}and{u}^3{v}^3=\frac{-8}{729}so{u}^{3}and{v}^{3}aresolution$$$$oftheequation{z}^2-\frac{16}{27}z-\frac{8}{729}=0...$$$$\mathrm{\Delta }={(-\frac{16}{27})}^2+\frac{32}{729}>0\Rightarrow z_1=\frac{\frac{16}{27}+\sqrt{\frac{{16}^2}{{27}^2}+\frac{32}{729}}}{2}$$$$z_2=...becontinued...$$$$$$$$$$

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