Help!!!!!

1. Let x and y be real numbers whose absolute values are different and that satisfy

$\begin{align*} \mathbf{ x^3 }&\mathbf{ = 20x + 7y } \\ \mathbf{ y^3 }&\mathbf{ = 7x + 20y } \end{align*}$

Find xy.

$\begin{array}{lrcl|c|c|} &&&& I. & II.\\ \hline (1) & x^3 &=& 20x + 7y \\ &&&& + & - \\ (2) & y^3 &=& 7x + 20y \\ \end{array}$

$\begin{array}{|rcll|} \hline (I.) & x^3+y^3 &=& 20x + 7y +7x + 20y \\ & x^3+y^3 &=& 27x + 27y \\ & x^3+y^3 &=& 27(x+y) \\ &&& x^3+y^3 = (x+y)(x^2-xy+y^2) \\ & (x+y)(x^2-xy+y^2) &=& 27(x+y) \\ & \mathbf{x^2-xy+y^2} &\mathbf{=}& \mathbf{27} \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline (II.) & x^3-y^3 &=& 20x + 7y -(7x + 20y) \\ & x^3-y^3 &=& 13x -13y \\ & x^3-y^3 &=& 13(x-y) \\ &&& x^3-y^3 = (x-y)(x^2+xy+y^2) \\ & (x-y)(x^2+xy+y^2) &=& 13(x-y) \\ & \mathbf{x^2+xy+y^2} &\mathbf{=}& \mathbf{13} \\ \hline \end{array}$

$\begin{array}{|lrcll|} \hline (1) & \mathbf{x^2-xy+y^2} &\mathbf{=}& \mathbf{27} \\ (2) & \mathbf{x^2+xy+y^2} &\mathbf{=}& \mathbf{13} \\ \hline (2)-(1): & x^2+xy+y^2 -(x^2-xy+y^2) &=& 13 - 27 \\ & x^2+xy+y^2 -x^2+xy-y^2 &=& -14 \\ & xy +xy &=& -14 \\ & 2xy &=& -14 \\ & \mathbf{xy} &\mathbf{=}& \mathbf{-7} \\ \hline \end{array}$

If x, y, and z are real numbers for which

$\begin{align*} \mathbf{x+y-z} &\mathbf{= -8,} \\ \mathbf{x-y+z} &\mathbf{= 18,\text{ and}} \\ \mathbf{-x+y+z} &\mathbf{= 30,} \\ \end{align*}$

then what is xyz?

$\begin{array}{|lrcl|} \hline (1) & x+y-z &=& -8 \\ (2) & x-y+z &=& 18 \\ (3) & -x+y+z&=& 30 \\ \\ \hline (1)+(2): & x+y-z + (x-y+z) &=& -8 + 18 \\ & 2x &=& 10 \\ & \mathbf{x} &\mathbf{=}& \mathbf{5} \\\\ \hline (1)+(3): & x+y-z + (-x+y+z) &=& -8 + 30 \\ & 2y &=& 22 \\ & \mathbf{y} &\mathbf{=}& \mathbf{11} \\\\ \hline (2)+(3): & x-y+z + (-x+y+z) &=& 18 + 30 \\ & 2z &=& 48 \\ & \mathbf{z} &\mathbf{=}& \mathbf{24} \\\\ \hline &xyz &=& 5\cdot 11 \cdot 24 \\ & \mathbf{xyz} &\mathbf{=}& \mathbf{1320} \\ \hline \end{array}$