Solve three equations with three unknowns Solve the system:
$$\begin{cases}a+b+c=6\\ab+ac+bc=11\\abc=6\end{cases}$$
The solution is:
$a=1,b=2,c=3$
How can I solve it?

Without Vieta's formulas, the polynomial can be derived the hard way by elimination. From the first and last equations:
$$
b+c=6-a \\
bc = 6 / a
$$
Substituting the above into the middle equation:
$$
11=a(b+c)+bc=a(6-a)+6/a \;\;\iff\;\; a^3-6 a^2+11a - 6 = 0
$$
By inspection, the latter equation has $a=1$ as a root, then factoring out $a-1$ gives:
$$a^3-6 a^2+11a - 6=(a-1)(a^2-5a+6)=(a-1)(a-2)(a-3)$$
Since the original system is symmetric in $a,b,c$ it follows that the solutions are $\{1,2,3\}$.
