It is the same with this: https://artofproblemsolving.com/community/u167643h2945588p26369678

no its not. this is $x^3$ not $x^2$

Yes, but the idea of the solution it is the same, instead of ${{x}^{8}}\ge x$ we have ${{x}^{27}}\ge x$ but this do not change anything.

It might be similar to the problem in #2 but there are some differences. Let's give a solution and see why. We have $x,y,z\ge 1$ since the roots are $\ge 0$. Rewriting the equations and adding yields $x^3+y^3+z^3+x+y+z=x^2+y^2+z^2+3$ Adding $x^3\ge x^2,y^3\ge y^2,z^3\ge z^2, x\ge 1,y\ge 1,z\ge 1$ we see that in all of them we must have equality. Hence $x=y=z=1$.

help me please i need a hint

Luchogamer7777 wrote: help me please i need a hint https://www.facebook.com/100069771979815/videos/3398176403840841

solution

indeed in all symetrical systems of equations x=y=z=etc that's the beauty of mathematics! George