# I have the following system of equations where the m's are known but a , b , c ,

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.
$\begin{array}{rl}a+b+c& ={m}_{0}\\ ax+by+cz& ={m}_{1}\\ a{x}^{2}+b{y}^{2}+c{z}^{2}& ={m}_{2}\\ a{x}^{3}+b{y}^{3}+c{z}^{3}& ={m}_{3}\\ a{x}^{4}+b{y}^{4}+c{z}^{4}& ={m}_{4}\\ a{x}^{5}+b{y}^{5}+c{z}^{5}& ={m}_{5}\end{array}$
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aniizl
Let $x,y,z$ be roots of ${t}^{3}+p{t}^{2}+qt+r=0$.
i.e.
${x}^{3}+p{x}^{2}+qx+r=0$ ----- (1)
${y}^{3}+p{y}^{2}+qy+r=0$ ----- (2)
${z}^{3}+p{z}^{2}+qz+r=0$ ----- (3)
Multiply the (1) by $a$, (2) by $b$ and (3) by $c$ and adding gives
${m}_{3}+p{m}_{2}+q{m}_{1}+r{m}_{0}=0$ ----- (4)
Multiply the (1) by $ax$, (2) by $by$ and (3) by $cz$ and adding gives
${m}_{4}+p{m}_{3}+q{m}_{2}+r{m}_{1}=0$ ----- (5)
Multiply the (1) by $a{x}^{2}$, (2) by $b{y}^{2}$ and (3) by $c{z}^{2}$ and adding gives
${m}_{5}+p{m}_{4}+q{m}_{3}+r{m}_{2}=0$ ----- (6)
Now (4), (5), (6) is a set of 3 linear equations in 3 variables ($p,q,r$) and can be solved easily.
This give us the cubic which $x,y,z$ satisfy (because we can find out that $p,q,r$ are) which can be solved using Cardano's Method, or more simply by the Trigonometric and Hyperbolic Method.
Once you know $x,y,z$ you can solve for $a,b,c$, as those become just linear equations.
codosse2e