Solve z-1/z=3y^3

$$\frac { z - 1 } { z } = 3 y ^ { 3 }$$

Solve for z

$$z=\frac{1}{1-3y^{3}}$$

$$y\neq \frac{3^{\frac{2}{3}}}{3}$$

Steps for Solving Linear Equation

Variable z cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by z.

$$z-1=3y^{3}z$$

Subtract 3y^{3}z from both sides.

$$z-1-3y^{3}z=0$$

Add 1 to both sides. Anything plus zero gives itself.

$$z-3y^{3}z=1$$

Combine all terms containing z.

$$\left(1-3y^{3}\right)z=1$$

Divide both sides by 1-3y^{3}.

$$\frac{\left(1-3y^{3}\right)z}{1-3y^{3}}=\frac{1}{1-3y^{3}}$$

Dividing by 1-3y^{3} undoes the multiplication by 1-3y^{3}.

$$z=\frac{1}{1-3y^{3}}$$

Variable z cannot be equal to 0.

$$z=\frac{1}{1-3y^{3}}\text{, }z\neq 0$$

Solve for y

$$y=\frac{3^{\frac{2}{3}}\sqrt[3]{\frac{z-1}{z}}}{3}$$

$$z\neq 0$$