How to find roots of a cubic through their sum and products? So I have a cubic polynomial and the sum and products of roots are as follows:
$x+y+z=-3$, 
 $xy+yz+xz=-6$,  $~xyz=18$
So i know this can be done through system of  simultaneous equations but dont know how, i know the answer but can someone show me the process?
BTW, the answers are $- 3,6^{1/2},-6^{1/2}$
 A: You can use Vieta's formulas. In particular, with an initial coefficient of $1$, i.e.,
$$P(s) = s^3 + a_2 s^2 + a_1 s + a_0 \tag{1}\label{eq1}$$
has $a_0 = -xyz = -18$, $a_1 = xy+yz+xz = -6$ and $a_2 = -(x + y + z) = 3$. Thus, you have
$$P(s) = s^2 + 3s^2 - 6s - 18 \tag{2}\label{eq2}$$
You could use this to solve for the individual roots using the Cubic function formula.
Update: As discussed in the question comments and the comments below, it's often easier & faster, especially for simpler type cubic equations (e.g., all coefficients are integers) to first try other methods to determine one root, e.g., by factoring by grouping (see How to Factor by Grouping for details) or using the Rational root theorem. If you do determine a root, you can then reduce your sets of equations to a quadratic, so you can then use the Quadratic formula to get the other $2$ roots. If you can't find an initial root, note the suggestion above of using the cubic function formula will always work.
A: Note that $$x^3-x^2(x_1+x_2+x_3)+x(x_1x_2+x_1x_3+x_2x_3)-x_1x_2x_3=x^3+px^2+qx+r$$
A: These can be solved simultaneously by progressively eliminating variables. Start with
$$
z = -3-x-y
$$
to get
$$
3y+y^2+3x+x^2+xy =6
\\ 3xy+x^2 y+y^2x =-18.
$$
Multiply the top through by $x$ and subtract the two equations to get
$$
x^3 +3x^2 = 6x + 18
$$
and we end up with the cubic equation anyways.
A: Assume $x_1$, $x_2$ and $x_3$ are the three roots, then they satisfy the equation
$$(x-x_1)(x-x_2)(x-x_3)=0\tag{1}$$
Expand the left-hand-side
$$x^3-(x_1+x_2+x_3)x^2+(x_1x_2+x_2x_3+x_3x_1)x-x_1x_2x_3=0$$
The given on the three roots, i.e. $x_1+x_2+x_3=-3$, 
 $x_1x_2+x_2x_3+x_3x_1=-6$ and $x_1x_2x_3=18$, just happen be the three coefficients, respectively. So, plug in their values to get
$$x^3+3x^2-6x-18=0\tag{2}$$
Since (1) and (2) are equivalent, We rewrite (2) below in the same product form as (1)
$$(x+3)(x-\sqrt{6})(x+\sqrt{6})=0\tag{3}$$
which can be verified straightforwardly by expanding. Then, compare (3) with (1), we are able to identify the three roots as $x_1-3$, $x_2=\sqrt{6}$ and $x_3=-\sqrt{6}$.
A: Consider the system
$$\begin{cases}x+y+z=a,\\xy+yz+zx=b,\\xyz=c\end{cases}$$ and try to solve it.
Multiplying the second by $z$,
$$xyz+z^2(x+y)=bz.$$
Then using the first and the second to eliminate $x,y$,
$$c+z^2(a-z)=bz$$ or
$$z^3-az^2+bz-c=0$$ and you are back at the cubic equation !
