Solve for x
\left\{\begin{matrix}x=\frac{z^{3}-1}{3y}\text{, }&y\neq 0\\x\in \mathrm{R}\text{, }&z=1\text{ and }y=0\end{matrix}\right.
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-3xy=1-z^{3}
Subtract z^{3} from both sides.
\left(-3y\right)x=1-z^{3}
The equation is in standard form.
\frac{\left(-3y\right)x}{-3y}=\frac{1-z^{3}}{-3y}
Divide both sides by -3y.
x=\frac{1-z^{3}}{-3y}
Dividing by -3y undoes the multiplication by -3y.
x=\frac{z^{3}-1}{3y}
Divide -z^{3}+1 by -3y.
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