Three-variable system of simultaneous equations

For a fixed number of variables and a fixed power $n$ the sum of powers $$x^n + y^n + z^n + ... + w^n$$ is a symmetric polynomial.

It is expressible in terms of elementary symmetric polynomials. The elementary symmetric polynomials for three variables are

• $e_1 = x + y + z$
• $e_2 = x y + x z + y z$
• $e_3 = x y z$

and your polynomials expressed in terms of them are

• $x + y + z = e_1$
• $x^2 + y^2 + z^2 = e_1^2 - 2 e_2$
• $x^3 + y^3 + z^3 = e_1^3 - 3(e_1 e_2 - e_3)$

Now we can find the values of $e_1,e_2,e_3$ evaluated at the given $x,y,z$:

$e_1 = 4$, $e_2 = 6$, $e_3 = 4$.

Now consider the polynomial $(t - x)(t - y)(t - z) = t^3 - e_1 t^2 + e_2 t - e_3 = t^3 - 4 t^2 + 6 t - 4$.

It has the solutions $t = 2, 1 + i$ and $1 - i$.

So now we can check if these are correct:

• $(2) + (1+i) + (1-i) = 4$
• $(2)^2 + (1+i)^2 + (1-i)^2 = 4 + 2i - 2i = 4$
• $(2)^3 + (1+i)^3 + (1-i)^3 = 8 + -2 + 2i -2 - 2i = 4$

You can write the elementary functions in terms of these sums of powers, thus getting the coefficients for a polynomial equation whose roots are $x$, $y$, and $z$. For instance, squaring the first equation and subtracting the second, you get the value of $xy+yz+xz$. Cubing the first equation and using what you now know, you get the value of $xyz$. In general you can use Newton's identities.

If you just want an answer, Wolfram Alpha makes short work of it, producing $(2,1+i,1-i)$ and all permutations of that. You expect six solutions from the product of the degrees. You can easily try x=y=z and y=z and find a contradiction, proving that the values of x,y, and z are all distinct. Then there is really only one solution, with permutations giving the six. If the solution is real, substitution will yield a sixth degree polynomial. If the solution is not all real, you must have one real value and two conjugate complex values, so set $y=a+bi, z=a-bi$ with $x, a$ and $b$ real. This gives

$\begin{align} &x+2a=4 \\ &x^2+2a^2-2b^2=4 \\ &x^3+2a^3-6ab^2=4 \end{align}$

which is actually easier than the real version. You can solve the second for $b^2$, insert it into the third along with $x=4-2a$ and get a cubic that will have $a=1$ as a solution.

I tried playing with the symmetric polynomials, but didn't find any slick approach.