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Is this the proper way?

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

FindInstance[{f, A > 0, B > 0, C > 0}, {A, B, C}, Integers, 5]
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  • $\begingroup$ a and A are different symbols, as are b and B. Next, AB is interpreted as a single vairbale called AB (and you probably want A*B). Finally your symbol (1/3) should be written out as (1/3) just as you wrote out (2/3). $\endgroup$ – bill s Jan 29 at 15:38
  • $\begingroup$ @bills thanks done $\endgroup$ – Dale Jan 29 at 15:40
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    $\begingroup$ There are still ABs occurring in f. Anyways, replacing AB by A B, Reduce[{f, A > 0, B > 0}, {A, B}, Integers] returns False, telling me that there are no solutions. $\endgroup$ – Henrik Schumacher Jan 29 at 16:09
  • $\begingroup$ You can simplify to f = A^3 - 6 B C^2 - 3 B^2 C + B^3 == 3 A (B^2 + 2 B C + C^2 + 2 A C) && A^3 + B^3 == C^3. $\endgroup$ – Somos Jan 29 at 20:52
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You can simplify to

f = A^3 - 6 B C^2 - 3 B^2 C + B^3 == 3 A (B^2 + 2 B C + C^2 + 2 A C) && A^3 + B^3 == C^3

but there are no positive solutions. Define

f[A_, B_] := With[{c = (A^3 + B^3)^(1/3)}, 
             A^3 - 6 B c^2 - 3 B^2 c + B^3 - 3 A (B^2 + 2 B c + c^2 + 2 A c)];

and Plot3D[ f[x, y], {x, 0, 2}, {y, 0, 2}] makes it obvious that f[x, y] < 0 for all positive values of x and y.

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