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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.

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It certainly depends on $a,b$. For example, take $a=b=0$, then the expression is $0 \in \mathbb{Z}$. Take $a=b=1$ and the expression is $2^{1/3}$ which is irrational. –  nullUser Jun 26 '12 at 18:51
@Bazinga I've added LaTeXed version of your formula. If this is what you meant, you can edit the post ant leave the LaTeX-ed version there. For more about writing math at this site see here or here. –  Martin Sleziak Jun 26 '12 at 18:51
Thanks @MartinSleziak –  Bazinga Jun 26 '12 at 18:53
Given any natural $a,b$, it will be an algebraic integer, therefore either integer or irrational. –  sdcvvc Jun 26 '12 at 18:57
Instead of saying "irrational, rational or integer", it may be more useful to say "irrational, non-integer rational, or integer". –  Dave L. Renfro Jun 26 '12 at 19:02

2 Answers 2

up vote 4 down vote accepted

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).

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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 $1$.

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