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

Question

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.

Answer 1

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).

Answer 2

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

Answer 3

See Johannes Blomer, How to denest Ramanujan's nested radicals, available here.