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.
 A: That problem has a unique solution, up to permutations of the variables. 
That happens since the power sums $p_k=x^k+y^k+z^k$ for $k=1,2,3$ give you the values of the elementary symmetric functions $e_1=x+y+z,e_2=xy+xz+yz,e_3=xyz$ through Newton's identities. Then $x,y,z$ can be identified with the roots of the polynomial:
$$ p(w) = w^3 - e_1 w^2 + e_2 w - e_3.$$
If you know in advance that $(x,y,z)=(-1,-3,5)$ works, then every solution is given by a permutation of $\{-1,-3,5\}$, since the coefficients of $p(w)$ are always the same, as well as its roots:
$$ p(w) = w^3 - w^2 -17w -15 = (w+1)(w+3)(w-5).$$
A: 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. 
