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Denest $\sqrt {\sqrt[3]{5}-\sqrt[3]{4}}$.

I have tried completing square by several method but all failed. Can anyone help me please? Thank you.

p.s. I'm a poor question-tagger.

Martin Sleziak
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JSCB
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4 Answers4

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This example is discussed in one of my prior posts, based on a polynomial-time denesting algorithm of Blomer. Using standard Galois theory of radical (Kummer) extensions, it is not difficult to prove a Denesting Structure Theorem, which implies that if a radical $\, r^{1/d} \,$ denests in any radical extension $\, F'$ of its base field $\, F$, then a suitable multiple $\, q b\, r \,$ of the radicand $\, r \,$ must already denest in the field $\, F' $ defined by the radicand. More precisely

Denesting Structure Theorem for Real Fields $\ $ Let $\, F \,$ be a real field and $\, F' = F(q_1^{1/d1},\ldots,q_k^{1/dk}) \,$ be a real radical extension of $\, F \,$ of degree $\, n. \,$ By $\, B = \{b_0,\ldots, b_{n-1}\}$ denote the standard basis of $\, F' \,$ over $\, F. \,$ If $\, r \,$ is in $\, F'$ and $\, d \,$ is a positive integer such that $\, r^{1/d} \,$ denests over $\, F \,$ using only real radicals, that is $\, r^{1/d} \in F(a_1^{1/t_1},\ldots,a_m^{1/t_m}) \,$ for positive integers $\, t_i \,$ and positive $\, a_i \in F,\,$ then there exists a nonzero $\, q \in F \,$ and $\, b \in B \,$ with $\, (q b r)^{1/d} \in F'.$

This implies that by multiplying the radicand by a $\, q \,$ in the base field $\, F \,$ and a power product $\, b \,=\, q_1^{e_1/d_1}\cdots q_k^{e_k/d_k} \,$ we can normalize any denesting so that it denests in the field defined by the radicand (then denesting reduces to solving for undetermined coefficients). For example

$$ \sqrt{\sqrt[3]5 - \sqrt[3]4} \,\,=\, \frac{1}3 (\sqrt[3]2 + \sqrt[3]{20} - \sqrt[3]{25})$$ normalises to $$ \sqrt{18\ (\sqrt[3]10 - 2)} \,\,=\, 2 + 2\ \sqrt[3]{10} - \sqrt[3]{10}^2 $$

An example with nontrivial $\,b$

$$ \sqrt{12 + 5\ \sqrt 6} \,\,=\, (\sqrt 2 + \sqrt 3)\ 6^{1/4} $$

normalises to

$$ \sqrt{\frac{1}3 \sqrt{6}\, (12 + 5\ \sqrt 6)} \,\,=\, 2 + \sqrt{6} $$

Here $\, F=\mathbb Q,\ F' = \mathbb Q(\sqrt 6),\ n=2,\ B = \{1,\sqrt 6\},\ d=2,\ q=1/3,\ b= \sqrt 6\,$.

The structure theorem also holds for complex fields except that we must assume that $\, F \,$ contains enough roots of unity (which may be computationally expensive in practice, to wit doubly-exponential complexity).

Bill Dubuque
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  • The "algorithm of Blomer" being, presumably, discussed in the link in my answer. – Gerry Myerson Sep 12 '12 at 05:56
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    @Gerry I think that paper is one of the references I cited in my answer here two years ago (based on one of my old sci.math posts). – Bill Dubuque Sep 12 '12 at 06:07
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    Is it perhaps a little tough to quote Galois theory of radical (Kummer) extensions when assisting someone struggling with an algebra-precalculus question? – Old John Sep 12 '12 at 21:55
  • @JohnSenior Please note that you don't need to understand any Galois theory to use the method in simple cases like this. It's analogous to the Risch structure theorem for integration in finite terms. Once you know the structure of the answer, finding the undetermined coefficients is simple algebra. There are no simpler approaches that work generally. – Bill Dubuque Sep 12 '12 at 22:22
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    That is true, but I was just empathising with an original poster being faced with an answer which is clearly at a much greater level of sophistication than the question. – Old John Sep 12 '12 at 22:31
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    @JohnSenior It occurs quite frequently in mathematics that elementary problems require nonelementary techniques for their solution. Here one might get lucky and correctly guess the structure of the answer. But mathematics is not based on pulling rabbits out of hats. When there are general methods that can be explained at an elementary level, we should strive to do so. – Bill Dubuque Sep 12 '12 at 22:51
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See Johannes Blomer, How to denest Ramanujan's nested radicals, available here.

Gerry Myerson
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There is a formula. I'm not too sure how to prove it, but I know that there is a formula where you can denest $$\sqrt{\sqrt[3]{\alpha}+\sqrt[3]{\beta}}$$Into$$\pm\frac {1}{\sqrt{f}}\left(-\frac {s^2\sqrt[3]{\alpha^2}}{2}+s\sqrt[3]{\alpha\beta}+\sqrt[3]{\beta^2}\right)$$ where $$f=\beta-s^3\alpha$$ and $s$ is a real number solution to $f(x)=x^4+4x^3+8\frac {\beta}{\alpha}x-4\frac {\beta}{\alpha}$

So in this case, $\alpha=5$ and $\beta=-4$. So $s=-2$ and $f=-4-(-2)^3\times 5=36$ Therefore, we have$$\pm\frac {1}{6}\left(-\frac {4\sqrt[3]{25}}{2}-2\sqrt[3]{-20}+\sqrt[3]{16}\right)=\pm\frac {1}{3}\left(-\sqrt[3]{25}+\sqrt[3]{20}+\sqrt[3]{2}\right)$$

Discard the negative value to get $\sqrt{\sqrt[3]{5}-\sqrt[3]{4}}=\frac {1}{3}\left(-\sqrt[3]{25}+\sqrt[3]{20}+\sqrt[3]{2}\right)$

Crescendo
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Frank
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I will give you a hint. Seek denesting which looks like this : $$ (\sqrt[3]{a} + \sqrt[3]{b} - \sqrt[3]{c})^2 = 9(\sqrt[3]{5}-\sqrt[3]{4}) $$

ivan
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