Solve $x+y+z=1; x^2+y^2+z^2=35; x^3+y^3+z^3=97$ It may be surprising that I can't get any analytical way of verifying that one of the solutions of $$x+y+z=1$$ $$x^2+y^2+z^2=35$$ $$x^3+y^3+z^3=97$$ is $x=-1, y=-3$ and $z=5$. Although it may be possible that I have to revisit my elementary arithmetic.

First note that
$$x^{2}+y^{2}+z^{2}=(x+y+z)^{2}-2(xy+yz+zx)=35$$ 
Now substituting $x+y+z=1$, we have $1^{2}-2(xy+yz+zx)=35$, and thus $xy+yz+zx=-17$. 
Then, $$x^{3}+y^{3}+z^{3}=(x+y+z)^{3}-3(x+y+z)(xy+yz+zx)+3xyz=97$$ and substituting with $1$  and our above result once again leads us to have  $xyz=15$. Now we need a product of $15$ and sum of $1$ for $(x,y,z)$, giving us $(-1, -3, 5)$ as a solution, amongst others. 
