(1)     a³+b³ = 72
(2)     a²-b² = 12

From (2)   b² = a²-12
(3)         b = ±Öa²-12

Substituting the negative solution in (1) and simplifying
gives a 6th degree equation with only imaginary solutions.
I will not deal with the imaginary solutions. So we will
only consider

(3)         b = Öa²-12

Substituting this positive square root and simplifying gives
this: 

   a4 - 4a3 - 12a2 + 192 = 0

which factors as

 (a-4)(a3-12a-48) = 0

This has rational solution a=4.  It also has one irrational
real solution and two imaginary solutions.  Substituting a=4
this in (3) gives b=2

Real rational solution:  a=4, b=2

I will not deal with the irrational real solution or the imaginary 
solutions.

Edwin