# How do you solve H=(J^3)/(CD) - (K^3)/D for J?

Oct 3, 2015

Rearrange using arithmetic operations to separate ${J}^{3}$ on one side of an equation, then take cube roots to get:

$J = \sqrt[3]{H C D + {K}^{3} C}$

#### Explanation:

First add ${K}^{3} / D$ to both sides to get:

$H + {K}^{3} / D = {J}^{3} / \left(C D\right)$

Multiply both sides by $C D$ to get:

$H C D + {K}^{3} C = {J}^{3}$

Take the cube root of both sides to get:

$J = \sqrt[3]{H C D + {K}^{3} C}$

This assumes Real arithmetic. If dealing with Complex numbers there would be two additional possible solutions:

$J = \omega \sqrt[3]{H C D + {K}^{3} C}$

$J = {\omega}^{2} \sqrt[3]{H C D + {K}^{3} C}$

where $\omega = - \frac{1}{2} + \frac{\sqrt{3}}{2} i$