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$\large \begin{cases} \dfrac{x+y}{1+z} & = & \dfrac{1-z+z^2}{x^2-xy+y^2} \\ \dfrac{x-y}{3-z} & = & \dfrac{z^2+3z+9}{x^2+xy+y^2} \end{cases}$

If $x,y,z$ satisfy the system of equations above, find the value of $z^3-y^3$.

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