How many real, rational and complex solutions has this system of equations? Let the system $\left\{\begin{aligned} 
      a+b+c &= 3\\ 
     a^2+b^2+c^2 &=5\\
a^3+b^3+c^3 & =12
    \end{aligned}\right. $
How many real, rational and complex solutions has it?
I read System of three variables of simultaneous equations and found $ e_1 = 3, e_2 = 2, e_3 = 1 $. Then I do not understand the reason but I get the polynomial $ t ^ 3-3t ^ 2 + 2t-1 $. The question is how many , not what are the solutions, so at this point I do not know what to do ...  or if I'm in the right way.  Can you help me please?
 A: It has 6 solutions
Here they are
$$a_1 = 2.32472$$
$$b_1 = 0.56228·i+0.337641$$
$$c_1 = 0.337641−0.56228·i$$
$$a_2 = 2.32472$$
$$b_ 2= 0.337641−0.56228·i$$
$$c_2 = 0.56228·i+0.337641$$
$$a_3 = 0.337641−0.56228·i$$
$$b_3 = 0.56228·i+0.337641$$
$$c_3 = 5.118e−18·i+2.32472$$
$$a_4 = 0.337641−0.56228·i$$
$$b_4 = 1.11022e−16·i+2.32472$$
$$c_4 = 0.56228·i+0.337641$$
$$a_5 = 0.56228·i+0.337641$$
$$b_5 = 0.337641−0.56228·i$$
$$c_5 = 1.69412e−18·i+2.32472$$
A: Note
$$ 3^2=(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca=5+2(ab+bc+ca). $$
from this, one has
$$ ab+bc+ca=2. $$
Also
\begin{eqnarray}
3^3&=&(a+b+c)^3=a^3+b^3+c^3+3a^2b+3ab^2+3a^2c+3ac^2+3b^2c+3bc^2+6abc\\
&=&12+3ab(a+b+c)+3bc(a+b+c)+3ca(a+b+c)-3abc\\
&=&12+3(ab+bc+ca)(a+b+c)-3abc\\
&=&30-3abc
\end{eqnarray}
from which, one has
$$ abc=1. $$
Thus $a,b$ and $c$ are the roots of the following equation
$$ t^3-3t^2+2t-1=0. \tag{1} $$
So the coefficients of (1) are
$$ a_3=1,a_2=-3,a_1=2,a_0=-1 $$
and hence the discriminant is
$$ \Delta=18a_3a_2a_1a_0-4a_2^3a_0+a_2^2a_1^2-4a_3a_1^3-27a_3^2a_0^2=-23<0$$
and hence (1) has one real root and two complex roots.
