Solving a peculiar system of equations I have the following system of equations where the $m$'s are known but $a, b, c, x, y, z$ are unknown. How does one go about solving this system? All the usual linear algebra tricks I know don't apply and I don't want to do it through tedious substitutions.
\begin{aligned}
a + b + c &= m_0 \\
ax + by + cz &= m_1 \\
ax^2 + by^2 + cz^2 &= m_2 \\
ax^3 + by^3 + cz^3 &= m_3 \\
ax^4 + by^4 + cz^4 &= m_4 \\
ax^5 + by^5 + cz^5  &= m_5
\end{aligned}
 A: @Aryabhata's solution in insightful, but I'll present the (tedious, but thoughtlessly mechanical) methodology I tend to use to approach such problems: resultants.
The resultant of two polynomials is a polynomial with one variable eliminated. Therefore, you can proceed to solve a polynomial system in roughly the same way you would with a linear system (in theory).
First, convert your equations to polynomials, $p_i$, so that the roots of these polynomials are the solutions to your equations.
$$\begin{eqnarray}
p_0 &=& a+b+c-m_0\\
p_1 &=& a x + b y + c z - m_1\\
p_2 &=& a x^2 + b y^2 + c z^2 - m_2\\
p_3 &=& a x^3 + b y^3 + c z^3 - m_3\\
p_4 &=& a x^4 + b y^4 + c z^4 - m_4\\
p_5 &=& a x^5 + b y^5 + c z^5 - m_5
\end{eqnarray}$$
Pick your favorite one (say, $p_0$), and use it to compute a bunch of resultants (using Mathematica's ${\tt Resultant[]}$ function) that eliminate one variable (say, $a$):
$$\begin{eqnarray}
r_0 &=& {\tt Resultant[} p_0, p_1, a {\tt ]} = -m_1 - b x - c x + m_0 x + b y + c z\\
r_1 &=& {\tt Resultant[} p_0, p_2, a {\tt ]} = -m_2 - b x^2 - c x^2 + m_0 x^2 + b y^2 + c z^2 \\
r_2 &=& {\tt Resultant[} p_0, p_3, a {\tt ]} = -m_3 - b x^3 - c x^3 + m_0 x^3 + b y^3 + c z^3\\
r_3 &=& {\tt Resultant[} p_0, p_4, a {\tt ]} = -m_4 - b x^4 - c x^4 + m_0 x^4 + b y^4 + c z^4\\
r_4 &=& {\tt Resultant[} p_0, p_5, a {\tt ]} = -m_5 - b x^5 - c x^5 + m_0 x^5 + b y^5 + c z^5\\
\end{eqnarray}$$
(Of course, the above amounts to simply solving $p_0 = 0$ for $a$ and substituting into the other $p_i$, since the system is linear in $a$. The same will be true when eliminating $b$ and $c$, so using resultants here is a little bit of overkill, but the process is mechanical enough to be about as easy to (not-)think about as, say, Gaussian elimination.)
Now, we use, say, $r_0$ to compute resultants that eliminate $b$, and so forth. For your special system, the polynomials at each step happen to have common factors that give rise to possible solutions; you might need to consider those separately, but they tend to represent "special case" scenarios. In the following, I base the next resultant step on the uncommon factors (indicated with upper-case variables).
$$\begin{eqnarray}
s_i &=& {\tt Resultant[} r_0, r_i, b {\tt ]} = S_i \cdot (x-y), \hspace{0.2in}i = 1, 2, 3, 4\\
t_i &=& {\tt Resultant[} S_0, S_i, c {\tt ]} = T_i \cdot (y-z)(z-x), \hspace{0.2in}i = 1, 2, 3 \\
u_i &=& {\tt Resultant[} T_0, T_i, x {\tt ]} = U_i \cdot (m_2-m_1 y - m_1 z + m_0 y z ), \hspace{0.2in}i = 1, 2 \\
v_1 &=& {\tt Resultant[} U_0, U_1, y {\tt ]}
\end{eqnarray}$$
The final polynomial, $v_1$, has just one variable, $z$.
$$\begin{eqnarray}v_1 &=& \left(m_2^2 - m_1 m_3 - m_1 m_2 z + m_0 m_3 z + m_1^2 z^2 - m_0 m_2 z^2\right)^4 \\ 
&& \left(-m_3^3 + 2 m_2 m_3 m_4 - m_1 m_4^2 - m_2^2 m_5 + m_1 m_3 m_5 \right.\\
&&\left. + z\left(m_2 m_3^2 - m_2^2 m_4 - m_1 m_3 m_4 + m_0 m_4^2 + m_1 m_2 m_5 - m_0 m_3 m_5 \right) \right. \\
&&\left. +z^2\left(- m_2^2 m_3 + m_1 m_3^2 + m_1 m_2 m_4 - m_0 m_3 m_4 - m_1^2 m_5 + m_0 m_2 m_5\right) \right. \\
&&\left. + z^3\left( m_2^3 - 2 m_1 m_2 m_3 + m_0 m_3^2 + 
   m_1^2 m_4 - m_0 m_2 m_4\right)\right)^2\end{eqnarray}$$
The first factor is another "special case", so that $z$, in general, should be a root of the cubic polynomial in the second factor. To get $y$, and $x$, and the rest, you could back-substitute into polynomials from earlier in the resultant chain (just as you would with a linear system), or you could just compute resultant chains that eliminate every variable except $y$, then every variable except $x$, etc.
It turns out that (as one might expect) your particular system is such that the "final" polynomials for $x$, $y$, and $z$ are equivalent; that is, $x$, $y$, and $z$ are roots of the cubic in the second factor of $v_1$ above. (This is, in fact, @Aryabhata's cubic.)
I'll note that resultants in general are computationally expensive (especially in symbolic form). Very often, the process can bog down completely in Mathematica, even for fairly modest systems, and/or the final polynomial can be of enormous degree. Sometimes, a bit of finesse in the order of elimination helps. My first attempt at finding (a chain of resultants that eliminated everything but) $a$ got boggy, but eliminating in the order $b$, $c$, $z$, $y$, $x$ gave me a cubic equation (and some "special case" factors) ... which is huge, so I won't present it here. A more-practical approach here might be to find $x$, $y$, and $z$ from the cubic polynomial, substitute these values back into three of the $p_i$, and solve the linear system for $a$, $b$, and $c$.
A: Let $\displaystyle x,y,z$ be roots of $\displaystyle t^3 + pt^2 + qt + r = 0$.
i.e. 
$\displaystyle x^3 + px^2 + qx + r = 0$        -----   (1)
$\displaystyle y^3 + py^2 + qy + r = 0$    -----  (2)
$\displaystyle z^3 + pz^2 + qz + r = 0$    ----- (3)
Multiply the (1) by $\displaystyle a$, (2) by $\displaystyle b$ and (3) by $\displaystyle c$ and adding gives
$\displaystyle m_3 + p m_2 + q m_1 + rm_0 = 0$    ----- (4)
Multiply the (1) by $\displaystyle ax$, (2) by $\displaystyle by$ and (3) by $\displaystyle cz$ and adding gives
$\displaystyle m_4 + pm_3 + qm_2 + rm_1 = 0$    ----- (5)
Multiply the (1) by $\displaystyle ax^2$, (2) by $\displaystyle by^2$ and (3) by $\displaystyle cz^2$ and adding gives
$\displaystyle m_5 + pm_4 + qm_3 + rm_2 = 0$    ----- (6)
Now (4), (5), (6) is a set of 3 linear equations in 3 variables ($\displaystyle p,q,r$) and can be solved easily.
This give us the cubic which $\displaystyle x,y,z$ satisfy (because we can find out that $\displaystyle p,q,r$ are) which can be solved using Cardano's Method, or more simply by the Trigonometric and Hyperbolic Method.
Once you know $\displaystyle x,y,z$ you can solve for $\displaystyle a,b,c$, as those become just linear equations.
A: HINT $\ $ Exploit the innate symmetry of the problem - here power sums as symmetric functions.  
The power sum $\rm\ \ \Sigma_n = a\ x^n + b\ y^n + c\ z^n\ \ $ satisfies the recursion
$$\rm \Sigma_{n+3} - (x+y+z)\ \Sigma_{n+2} + (xy+yz+zx)\ \Sigma_{n+1} - xyz\ \Sigma_n\ =\ 0 $$
For $\rm\ n = 0,1,2\ $ the above yields a linear system for $\rm\: \ x+y+z,\ \ xy+yz+zx,\ \ xyz\ ,\:$ which are the coefficients of the cubic whose roots are $\rm\ x, y, z$
REMARK $\ $ As I mentioned in an answer to a similar question, readers familiar with the  theory of Lucas-Lehmer sequences may recognize this recursion as one of many useful identities satisfied by sums of powers.  Such identities arise in many diverse contexts. Here is an interesting example: Capdegelle's work on FLT (Fermat's Last Theorem) employed the power sum recursion $\rm\ F_{n+3} + S_2\: F_{n+2} + S_1\: F_{n+1} + S_0\: F_n = 0\ $ where $\rm\ F_n = X^n + Y^n - Z^n\ $ is the Fermat polynomial and the polynomials $\rm\ S_k\ $ are the elementary symmetric polynomials $\rm\ S_0 = -XYZ,$ $\rm\ S_1 = XY + XZ + YZ,\ $ $\rm S_2 = -(X+Y+Z)\:.\ $ 
