Prove that $(5^{\frac 12} + 2)^{\frac 13}- (5^{\frac 12} - 2)^{\frac 13}$ is an integer and find its value I had proceed this question by taking
$$x =(5^{\frac 12} + 2)^{\frac 13}- (5^{\frac 12} - 2)^{\frac 13}$$
Then
$$x + (5^{\frac 12} -2)^{\frac 13} =(5^{\frac 12} + 2)^{\frac 13}$$
And then cubing both sides and then solving for $x$ by using Cardano's method but I got the same equation at last that is
$$x =(5^{\frac 12} + 2)^{\frac 13}- (5^{\frac 12} - 2)^{\frac 13}$$
Now I don't know how to solve this

Some naive bounding gives
$(\sqrt 5 + 2)^{\frac 13} - (\sqrt 5 - 2)^{\frac 13} < (\sqrt 5 + 2)^{\frac 13} < (3+2)^{\frac 13} < 8^\frac 13 = 2$.
It is also easy to show that this quantity is positive, so if it is an integer then it must be $1$.
Some Galois theory shows that if $(\sqrt 5 + 2)^{\frac 13} - (\sqrt 5 - 2)^{\frac 13}$ really is an integer then $(\sqrt 5 + 2)$ is a cube in the ring of integers of $\Bbb Q(\sqrt 5)$, so $(\sqrt 5 + 2) = ((a + b\sqrt 5)/2)^3$ where $a$ and $b$ are some integers with $a \equiv b \pmod 2$. Conversely, if this is the case then one can quickly show $(\sqrt 5 + 2)^{\frac 13} - (\sqrt 5 - 2)^{\frac 13} = (a + b\sqrt 5)/2 - (-a + b\sqrt 5)/2 = a$.  
Since we know this can only be $1$, we must have $a=1$. Then writing $(1 + b\sqrt 5)^3 = 16 + 8\sqrt 5$, we get the equations $1+15b^2 = 16$ and $3b+5b^3 = 8$. The first one gives $b^2=1$ and then the second one gives $8b = 8$ so $b=1$, which is compatible with $b^2=1$, so everything works out well.
If you don't know Galois theory, assuming that the cube root of $\sqrt 5 +2$ has such a nice form sounds like a gamble, but it is a gamble that is secretly rigged in our favour. 
