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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.

miradulo
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Hassan Muhammad
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  • To verify that $(-1,-3,5)$ is a solution, you plug those numbers in each equation. So $-1 + (-3) + 5 = 1$, $(-1)^2 + (-3)^2 + 5^2 = 1 + 9 + 25 = 35$, and $(-1)^3 + (-3)^3 + 5^3 = -1 - 27 + 125 = 97$. Each equation comes out true and so $(-1,-3,5)$ is a valid solution to the system of equations. – Decaf-Math Jul 13 '15 at 15:43

2 Answers2

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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:

$$ p(w) = w^3 - w^2 -17w -15 = (w+1)(w+3)(w-5).$$

Jack D'Aurizio
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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.

miradulo
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