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#