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

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Answer 1 · 2015-10-03T21:46:23

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

$J = root(3)(HCD+K^3C)$

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

$H + K^3/D = J^3/(CD)$

Multiply both sides by $CD$ to get:

$HCD + K^3C = J^3$

Take the cube root of both sides to get:

$J = root(3)(HCD+K^3C)$

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

$J = omega root(3)(HCD+K^3C)$

$J = omega^2 root(3)(HCD+K^3C)$

where $omega = -1/2 + sqrt(3)/2i$

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