Firstly, let 4b^1/2 = B, so that the equation we must solve is (A+B)^1/3 + (A-B)^1/3 = 3.

Multiply and divide by the conjugate (A+B)^1/3 - (A - B)^1/3:

(A+B)^2/3 - (A-B)^2/3 = 3(A+B)^1/3 - 3(A - B)^1/3

[4AB]^1/3 = 3(A+B)^1/3 - 3(A - B)^1/3

(A+B)^1/3 - (A - B)^1/3 = [4AB]^1/3/3.

So that gives us two equations to work with to solve for A in terms of B (I know they're the same equation technically, it's just a trick):

• (A + B)^1/3 + (A - B)^1/3 = 3
• (A + B)^1/3 - (A - B)^1/3 = [4AB]^1/3/3

Add the equations and subtract the equations:

2(A + B)^1/3 = 3 + [4AB]^1/3/3

2(A - B)^1/3 = 3 - [4AB]^1/3/3

Cube the equations.

8(A + B) = 27 + 3[4AB]^1/3 + [4AB]^2/3/3 + 4AB/27

8(A - B) = 27 - 3[4AB]^1/3 + [4AB]^2/3/3 - 4AB/27

Add the equations:

16A = 54 + 2/3 [4AB]^2/3

[4AB]^2/3 = (24A - 81)

16A²B² = (24A - 81)³

256a²b = (24a - 81)³

This gives you b as a function of a, so the last step is determining when (24a - 81)³/(16a)² is an integer.

There may be errors in my algebra, so double check it.