Solving $n$ unknowns with $n$ independent equations Is it always possible to solve $n$ independent equations with $n$ unknowns?
Or is it possible to solve the following 3 equations with 3 unknowns?
$$x + y + z = a$$
$$x^2 + y^2 + z^2 = b$$
$$x^3 + y^3 + z^3 = c$$
 A: This is a classical example of a system of polynomial equations, which is often given
as an application of Groebner bases. First we need to say, where $a,b,c$ belong to, i.e.,
over which field we want to consider the equations. Usually, we have the field of complex or real numbers, but also Diophantine equations over the rational numbers or integers are possible. Then we can compute a Groebner basis, depending on the given $a,b,c\in K$.
For example, for $(a,b,c)=(1,1,1)$ we have the following solutions:
$$
(x,y)=(1,0),(0,0),(0,1),
$$
and $z=1-x-y$.
In general, we can express, say, $x$ and $z$ in terms of $y$, namely $z=a-x-y$ and $x$ a root of the quadratic equation
$$
2x^2 + 2xy + 2y^2 - (2a)x - (2a)y + (a^2 - b)=0.
$$
Then $y$ has to satisfy a cubic polynomial equation, namely
$$
6y^3 - (6a)y^2 + (3a^2 - 3b)y - (a^3 - 3ab + 2c)=0.
$$
Over the complex numbers this equation has $3$ solutions.
A: This problem can be solved via finding roots of polynomial of $n$-th degree.
$2$ links are helpful here:


*

*Newton's Identities;

*Vieta's formulas.  


Denote (according to $1$st link) 
$$
p_1 = x+y+z=a;\\
p_2 = x^2+y^2+z^2=b;\\
p_3 = x^3+y^3+z^3-c;
$$
and
$$
e_1 = x+y+z;\\
e_2 = xy+xz+yz;\\
e_3 = xyz.
$$
Then
$$
e_1=p_1=a;\\
2e_2 = e_1p_1-p_2=a^2-b;\\
3e_3 = e_2p_1-e_1p_2+p_3 = \dfrac{a^2-b}{2}a-ab+c;
$$
so
$$
e_1 = a;\\
e_2 = \dfrac{a^2-b}{2};\\
e_3 = \dfrac{a^3-3ab+2c}{6}.
$$
Then (according to $2$nd link) $x,y,z$ are three roots of polynomial
$$
p(x) = x^3-e_1x^2+e_2x-e_3.
$$

Example ($a=18, b=134, c=1152$):
$$
x+y+z=18;\\
x^2+y^2+z^2=134;\\
x^3+y^3+z^3=1152;
$$
then
$$
e_1=18, \quad e_2=95, \quad e_3=150;
$$
so $x,y,z$ are the roots of polynomial 
$$
p(x) = x^3-18x^2+95x-150.
$$
$(x=3,y=5,z=10)$ and their permutations.
A: Consider $x$ , $y$ and $z$ as the roots of the equation :
$$(X-x)(X-y)(X-z)=X^3+AX^2+BX+C=0$$
$A=-(x+y+z)=-a$
$B=xy+yz+zx=\frac{1}{2}\left((x+y+z)^2-(x^2+y^2+z^2)\right)=\frac{1}{2}(a^2-b)$
$C=-xyz=-\frac{1}{6}\left( (x+y+z)^3-3(x+y+z)(x^2+y^2+z^2)+2(x^3+y^3+z^3) \right)=-\frac{1}{6}(a^3-3ab+2c)$
$x$ , $y$ , $z$ are the roots of the equation :
$$X^3-aX^2+\frac{1}{2}(a^2-b)X-\frac{1}{6}(a^3-3ab+2c)=0$$
