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Question Number 6809 by Tawakalitu. last updated on 28/Jul/16

$$If{x}^{\frac{1}{3}}+y^{\frac{1}{3}}+z^{\frac{1}{3}}=0,Thenpovethat$$$${(x+y+z)}^3=3xyz$$

Commented by Yozzii last updated on 29/Jul/16

$$x^{1/3}+y^{1/3}+z^{1/3}=0$$$$Letx=u^3,y=w^3,z=v^3.$$$$\therefore u+w+v=0$$$$Usingtheidentity{u}^3+w^3+v^3=(u+w+v)(u^2+w^2+v^2-uv-vw-wu)+3uwv$$$$andu+w+v=0,$$$$3uvw=u^3+w^3+v^3$$$$or3{\left(xyz\right)}^{1/3}=x+y+z$$$$\Rightarrow {(x+y+z)}^3=27xyz$$$$-----------------------$$$$Letx=8,y=27,z=-125asanexample.$$$$\Rightarrow x^{1/3}+y^{1/3}+z^{1/3}=2+3-5=0.$$$$\therefore {(x+y+z)}^3=-729000$$$$And3xyz=3\times 8\times 27\times (-125)=-81000\ne -729000$$$$\Rightarrow {(x+y+z)}^3\ne 3xyzgenerallyif{x}^{1/3}+y^{1/3}+z^{1/3}=0.$$$$But,27xyz=27\times 8\times 27\times (-125)=-729000$$$$\Rightarrow {(x+y+z)}^3=27xyz.$$

Commented by Tawakalitu. last updated on 29/Jul/16

$$Thankyouverymuch.$$

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