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I read in a book (A Synopsis of Elementary Results in Pure and Applied Mathematics) that the condition to simplify the expression $\sqrt[3]{a+\sqrt{b}}$ is that $a^2-b$ must be a perfect cube.

For example $\sqrt[3]{10+6\sqrt{3}}$ where $a^2-b =(10)^2-(6 \sqrt{3})^2=100-108=-8$ and $\sqrt[3]{-8} = -2$ So the condition is satisfied and $\sqrt[3]{\sqrt{3}+1}^3=\sqrt{3}+1$.

But the example $\sqrt[3]{11+\sqrt{57}}$ where $a^2-b = (11)^2-57=121-57=64$ and $\sqrt[3]{64}=4$ so the condition is satisfied.

But I can’t simplify this expression. Please help us to solve this problem. Note: this situation we face it in many examples

Caleb Stanford
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Hashem
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  • It's very hard to understand what the question here is... – barak manos May 11 '14 at 19:49
  • The book (which can be found online) states the condition as necessary: "No simplification is effected unless $a^2-b$ is a perfect cube." It doesn't say anything about the condition being sufficient, and apparently it isn't. – Barry Cipra May 12 '14 at 01:16

4 Answers4

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I imagine you want to write, with integer $u$ and $v$,

$$11+\sqrt{57}=(u+\sqrt{v})^3$$

So that the cube root would simplify.

Hence

$$u^3+3u^2\sqrt{v}+3uv+v\sqrt{v}=11+\sqrt{57}$$

So you would need

$$\begin{eqnarray} u(u^2+3v)&=&11\\ v(3u^2+v)^2&=&57 \end{eqnarray}$$

But $57=3 \cdot 19$ has no square factor, so $3u^2+v=\pm1$, and $v=57$. But then the first equation can't hold, since $u^2+3v\geq 3\cdot57=171$.

So it's in fact not possible to simplify this way.

Jean-Claude Arbaut
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Here’s the story: The real quadratic number field $\mathbb Q(\sqrt{57}\,)$ fortunately has class number $1$ (I looked it up on the internet), so that it is relatively easy to do arithmetic there. The fundamental unit is $\varepsilon=151+20\sqrt{57}$, which has norm $+1$, so that $1/\varepsilon$ is $151-20\sqrt{57}$. The other thing we need to do is know how the prime $2$ behaves there. Since $57\equiv1\pmod8$, the prime $2$ splits, indeed $-2=zz'$, where $z=(7+\sqrt{57}\,)/2$, and $z'$ is the same thing but with a minus sign.

Now here’s how to factor $11+\sqrt{57}$: $$ 11+\sqrt{57}=2\frac{11+\sqrt{57}}2=-zz'\frac{11+\sqrt{57}}2\,, $$ but, setting $(11+\sqrt{57}\,)/2=w$, we compute that $w/z^4=151-20\sqrt{57}=1/\varepsilon$. Thus the factorization of our original number is $$ 11+\sqrt{57}=-\varepsilon^{-1}z^5z'\,, $$ which is definitely not a cube in our field. In case you look instead at $11-\sqrt{57}$, it will get expressed as $-\varepsilon z{z'}^5$, still not a cube.

Lubin
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It is not a sufficient condition (I don't know if it's necessary). Not all expressions of the form $\sqrt[3]{a+\sqrt{b}}$, satisfying the condition that $a^2-b$ is a perfect cube, can be simplified.

user26486
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  • Can you give a reason why $a^2-b$ is a perfect square does not imply $a+\sqrt{b}$ can be written as a perfect cube? – Sandeep Silwal May 11 '14 at 19:56
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    @SandeepSilwal I do not know what you mean. I said, "not all expressions", an example being the OP's example. –  May 11 '14 at 19:57
  • you can see the book (A Synopsis of Elementary Results in Pure and Applied Mathematics if you like – Hashem May 11 '14 at 20:01
  • It is a necessary condition because $N(x+y\sqrt{b}) = x^2 - b y^2$ is a multiplicative function on ${\mathbb Z} + {\mathbb Z} \sqrt{b}$. But, as this example shows, not a sufficient condition. – Robert Israel May 11 '14 at 20:05
  • What the book probably means is something like the following. If a number of the form $a+b\sqrt n$ is a cube of another number of the same form, i.e. $(u+v\sqrt n)^3=a+b\sqrt n$, then this implies that we also have $(u-v\sqrt n)^3=a-b\sqrt n$. Multiplying these two equations we get $$a^2-b=(a-b\sqrt n)(a+b\sqrt n)=\cdots=(u^2-nv^2)^3.$$ In this sense the condition is necessary. I cannot tell to what extent this can be reversed, so I don't know if the condition is sufficient in any sense. All $a,b,u,v,n$ above are integers. – Jyrki Lahtonen May 11 '14 at 20:07
  • IMHO this would be better as a comment as it does not really answer the question. I'm not the downvoter in case you were wondering :-/ – Jyrki Lahtonen May 11 '14 at 21:04
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I think we want a "-" sign not a "+" sign.

Squaring under the radical gives 64. Then the square root and a cube root gives 2.

user1475
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