Will the following expression be irrational, rational or integer? Will the following expression be irrational, rational or integer?
$$\sqrt[3]{\sqrt a +b} - \sqrt[3]{\sqrt a -b}$$
where $a$ = $52$ and $b$ = $5$ .
By intuition, I think this will be an integer. 
 A: The expression quickly brings to my mind the Cardano formulas for a (depressed) cubic. Which suggested the following:
We can rewrite as
$$\sqrt[3]{b+\sqrt{a}} + \sqrt[3]{b - \sqrt{a}}.$$
This is a root of the cubic $y^3 + 3py + q=0$, where
$b = \frac{-q}{2}$ and $a=\frac{q^2+4p^3}{4}$.
With $b=5$, we have $q=-10$, so
$$52 = a = \frac{100+4p^3}{4}.$$
This gives $4p^3 = 208-100 = 108$, so $p^3 = 27$, or $p=3$.
Thus, the expression at hand is a root of the cubic
$$y^3 + 9y -10 = 0.$$
This cubic has an obvious integer root $y=1$ (or you could use the rational root test to see if it has any rational roots if this is not obvious), which after factoring gives
$$y^3 + 9y - 10 = (y-1)(y^2 + y+10).$$
The quadratic is irreducible over $\mathbb{R}$, so its roots are complex.
Assuming you mean the real cubic roots of the real numbers, since our number is real, it must equal $1$.
A: Let's use the identity $(\alpha + \beta)^3 = \alpha^3+\beta^3+3\alpha\beta(\alpha+\beta)$
Set $\alpha =\sqrt[3]{b+\sqrt{a}} \text{ and } \beta= \sqrt[3]{b - \sqrt{a}} \text { and } \alpha+\beta=x$
We know that $\alpha\beta = \sqrt[3]{b+\sqrt{a}} \times \sqrt[3]{b - \sqrt{a}} = \sqrt[3]{b^2-a} = \sqrt[3]{25-52} = -3$
So $x^3= b+\sqrt a + b-\sqrt a - 9x = 2b-9x = 10-9x$
You know that you are looking for a real answer, because $a$ is positive and the two cube roots are therefore cube roots of real numbers. Arturo has factorised the cubic, but it is easy to see that 1 is a root (integer roots must be divisors of 10).
