Solving a system of equations $(x+y-z=1), (x^2+y^2-z^2=5-2xy), (x^3+y^3-z^3=43-3xy)$

Question

I have to solve this system of equations \begin{cases} x+y-z=1 \\ x^2+y^2-z^2=5-2xy \\x^3+y^3-z^3=43-3xy \end{cases}

From the third equation I have done this to manipulate it:

$$ x^3+y^3-z^3=43-3xy \implies (x+y)(x^2-xy+y^2)-z^3=43-3xy\implies (x+y)((x+y)^2-3xy) -z^3=43-3xy$$

Notice from the first equation $x+y=z+1$ and $(x+y)^2 = z^2+5$

So the above equation reduces to $z^2+5z-38-3xyz=0$

How do I get rid of the $xyz$?

Answer 1

$$ x^3+y^3-z^3=43-3xy \implies ...\implies (x+y)((x+y)^2-3xy) -z^3=43-3xy$$

From the first equation: $\color{blue}{x+y}=\color{green}{z+1}$

$$(\color{blue}{x+y})^2=z^2+5$$

$$(\color{green}{z+1})^2=z^2+5$$

$$\vdots$$

It's easy from this point

Answer 2

$$\begin{cases} x+y-z=1 \\ x^2+y^2-z^2=5-2xy \\x^3+y^3-z^3=43-3xy \end{cases}\Rightarrow z^2+5=z^2+2z+1\Rightarrow z=2$$ Because of $x^3+y^3=(x+y)(x^2-xy+y^2$ it follows $$\begin{cases} x+y=3 \\ x^2+y^2=17 \end{cases}\Rightarrow (x,y)=(-1,4),(4,-1)$$