Solve y^3-1/yz=68/3

$$y ^ { 3 } - \frac { 1 } { y z } = \frac { 68 } { 3 }$$

Solve for z

$$z=\frac{3}{y\left(3y^{3}-68\right)}$$

$$y\neq \frac{\sqrt[3]{68}\times 3^{\frac{2}{3}}}{3}\text{ and }y\neq 0$$

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 3yz, the least common multiple of yz,3.

$$3yzy^{3}-3=68yz$$

To multiply powers of the same base, add their exponents. Add 1 and 3 to get 4.

$$3y^{4}z-3=68yz$$

Subtract 68yz from both sides.

$$3y^{4}z-3-68yz=0$$

Add 3 to both sides. Anything plus zero gives itself.

$$3y^{4}z-68yz=3$$

Combine all terms containing z.

$$\left(3y^{4}-68y\right)z=3$$

Divide both sides by 3y^{4}-68y.

$$\frac{\left(3y^{4}-68y\right)z}{3y^{4}-68y}=\frac{3}{3y^{4}-68y}$$

Dividing by 3y^{4}-68y undoes the multiplication by 3y^{4}-68y.

$$z=\frac{3}{3y^{4}-68y}$$

Divide 3 by 3y^{4}-68y.

$$z=\frac{3}{y\left(3y^{3}-68\right)}$$

Variable z cannot be equal to 0.

$$z=\frac{3}{y\left(3y^{3}-68\right)}\text{, }z\neq 0$$