<s>[INST] Solving a system of equations $(x+y-z=1), (x^2+y^2-z^2=5-2xy), (x^3+y^3-z^3=43-3xy)$ 

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 $xyz$?
 [/INST] $$ x^3+y^3-z^3=43-3xy \implies ...\implies (x+y)((x+y)^2-3xy) -z^3=43-3xy$$
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
</s>