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
 A: As you say let
$$x = \sqrt[3]{\sqrt 5+2} - \sqrt[3]{\sqrt 5-2}.$$
If you now take the cube you get
$$x^3 = 4-3x.$$
The polynomial factors into
$$x^3+3x-4=(x-1)(x^2+x+4).$$
So $x=1$ is the only real solution to the equation
$$x^3+3x-4=0.$$
A: Let $$x=\sqrt[3]{\sqrt5+2}-\sqrt[3]{\sqrt5-2},$$ 
raise it to the third power:
$$\begin{align}
x^3&=\sqrt5+2-3\sqrt[3]{\sqrt5+2}^2\sqrt[3]{\sqrt5-2}
+3\sqrt[3]{\sqrt5+2}\sqrt[3]{\sqrt5-2}^2-(\sqrt5-2)\\
&=4-3\cdot\sqrt[3]{\sqrt5+2}\sqrt[3]{\sqrt5-2}\cdot\bigl(\sqrt[3]{\sqrt5+2}-\sqrt[3]{\sqrt5-2}\bigr)\\
&=4-3\cdot\sqrt[3]{5-4}\cdot x\\
&=4-3\cdot1\cdot x.
\end{align}
$$
Now show that $x=1$ is the only solution of $x^3+3x-4=0$.
A: Let $a=\sqrt[3]{\sqrt{5}+2}$, $b=\sqrt[3]{\sqrt{5}-2}$ and $x=a-b$. Then,
$$x^3=\left(a-b\right)^3=a^3-3a^2b+3ab^2-b^3=a^3-b^3-3ab\left(a-b\right)\\=\sqrt{5}+2-\left(\sqrt{5}-2\right)-3x\sqrt[3]{\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)}=4-3x$$
Then the answer is the real root of the equation $x^3+3x-4=0$,
$$x^3-3x-4=0\\\left(x-1\right)\left(x^2+x+4\right)=0\\\because x^2+x+4>0\\ \therefore x=1$$ We get $\sqrt[3]{\sqrt{5}+2}-\sqrt[3]{\sqrt{5}-2}=1$
A: Assume that
$$\sqrt5\pm2=(a\sqrt5\pm b)^3=(5a^3+3ab^2)\sqrt5\pm(15a^2b+b^3).$$
We observe that $5+3=8$ and $15+1=2\cdot8$ and conclude
$$\sqrt5\pm2=\left(\frac{\sqrt5\pm1}2\right)^3.$$
Hence $1$.
A: 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. 
