CPhill's answer is, in the words of one of my instructors, elegant.
That won't be said for mine; more like a brute force attempt:
To make the typing easier,
let A = (9 + 4√5)^(1/3) ---> A³ = 9 + 4√5
let B = (9 - 4√5)^(1/3) ---> B³ = 9 - 4√5
Let X = A + B So, now: X = (9 + 4√5)^(1/3) + (9 - 4√5)^(1/3) (the problem)
Also: A·B = [ (9 + 4√5)^(1/3) ] · [ (9 - 4√5)^(1/3) ]
= [ (9 + 4√5) · (9 - 4√5) ] ^(1/3)
= [ 81 - 16 · 5 ] ^(1/3)
= [ 1 ] ^(1/3)
= 1
Since A·B = 1,
A²·B = A(AB) = A(1) = A = (9 + 4√5)^(1/3)
A·B² = (AB)B = (1)B = B = (9 - 4√5)^(1/3)
Sinc X = A + B,
X³ = (A + B)³ = A³ + 3A²·B + 3A·B² + B³
X³ = (9 + 4√5) + [ 3(9 + 4√5)^(1/3) ] + [ 3(9 - 4√5)^(1/3) ] + ( 9 - 4√5 )
Rearranging:
X³ = 18 + 3[ (9 + 4√5)^(1/3) + (9 - 4√5)^(1/3) ]
X³ = 18 + 3[ A + B ]
But, since A + B = X
X³ = 18 + X
X³ - X - 18 = 0
Factoring:
(X - 3)(X² + 3X + 6) = 0
So: X = 3 or X = [-3 ± i√(15) ] / 2
Since the answer is a pure real number, the answer is 3!