Solving the system $x+y+z=1$, $x^2+y^2+z^2=2$, $xyz-xy-xz-yz=3$

Question

$$\begin{align} x+y+z&=1 \\ x^2+y^2+z^2&=2 \\ xyz-xy-xz-yz&=3 \end{align}$$

My Attempt:

$(x+y+z)^2=x^2+2xy+2xz+y^2+2yz+z^2=1$
$xyz-xy-xz-yz=3$
$xy+xz+yz+3=xyz$
$xyz-3=xy+xz+yz$
$1=x^2+2xy+2xz+y^2+2yz+z^2$
$1=2+2xy+2xz+2yz$
$1=2+2xyz-6$
$5=2xyz$
$\frac{5}{2}=xyz$
$xyz-3=xy+xz+yz$
$-\frac{1}{2}=xy+xz+yz$

$f(x)=f(y)=f(z)=0$
$f(t)=(t-x)(t-y)(t-z)$
$=t^3-(x+y+z)t^2+(xy+xz+yz)t-xyz$
$=t^3-t^2-\frac{t}{2}-\frac{5}{2}$
all roots of f(x) are the solutions
Are there any ways to get solutions other than the cubic formula?

Answer 1

You have proved that $x,y,z$ are roots of $2t^3-2t^2-t-5=0$.

None of the possible rational roots works, so you have to assume irrational roots and use either numerical methods or the cubic formula.

We can prove that only one root is real. Descartes' Rule of Signs guarantees a positive root $r$, and dividing out the factor $t-r$ must give

$\color{blue}{t^2-(1-r)t+5/(2r)=0}$

from the sum and product of the remaining roots. For those remaining roots to also be real we must then have

$(1-r)^2-(10/r)\ge0$

$(r^2-2r+1)-(10/r)\ge0$

$r^3-2r^2+r-10\ge0$ (since $r>0$)

But also $2r^3-2r-r-5=0$, so

$2(r^3-2r^2+r-10)-(2r^3-2r^2-r-5)$

$=-(2r^2-6r+15)\ge0.$

Then $2r^2-6r+15$ has a negative discriminant itself and so is everywhere positive, a contradiction. Therefore the positive root $r$ guaranteed by the Rule of Signs is the only real one. So we use the cubic formula or a numerical method to find this root and then solve the blue equation for the complex conjugates. The three roots are then the roots for $x,y,z$ in any order.