solve for x,y, and z:
x+y = 3z
x^2 + y^2 = 5z
x^3 + y^3 = 9z
Hello, jmedsy!
I think I've solved it . . . I'll get you started.
Solve for $\displaystyle x,y,z\!:\;\;\begin{array}{cccc}x+y &=& 3z & [1] \\ x^2 + y^2 &=& 5z & [2] \\ x^3 + y^3 &=& 9z & [3] \end{array}$
Note that $\displaystyle x=y=z=0$ is a solution.
In the following work, all the variables are nonzero.
Square [1]: .$\displaystyle (x+y) \:=\:(3z)^2 \quad\Rightarrow\quad x^2 + 2xy + y^2 \:=\:9z^2$
. . . $\displaystyle \underbrace{x^2+y^2}_{\text{This is }5z} + 2xy \:=\:9z^2 \quad\Rightarrow\quad 5z + 2xy \:=\:9z^2 \quad\Rightarrow\quad xy \:=\:\frac{9z^2-5z}{2}$ .[1]
Cube [1]: .$\displaystyle (x+y)^3 \:=\:(3z)^3 \quad\Rightarrow\quad x^3 + 3x^2y + 2xy^2 + y^3 \:=\:27z^3 $
. . . $\displaystyle \underbrace{x^3+y^3}_{\text{This is }9z} + 3xy\underbrace{(x+y)}_{\text{This is }3x} \:=\:27z^2 \quad\Rightarrow\quad 9z + 9xyz \:=\:27z^3 \quad\Rightarrow\quad xy \:=\:3z^2-1$ .[2]
Equate [1] and [2]: .$\displaystyle \frac{9z^2-5z}{2} \:=\:3z^2-1 \quad\Rightarrow\quad 3z^2 - 5z + 2 \:=\:0$
Factor: .$\displaystyle (z-1)(3z-2) \:=\:0 \quad\Rightarrow\quad z \:=\:1,\:\tfrac{2}{3}$
Your turn . . .