Is it possible to represent $$ \sqrt[3] {7\sqrt[3]{20}-1} =\sqrt[3]{A}+\sqrt[3]{B}+\sqrt[3]{C}$$ with rational $A,\,B,$ and $C?$
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Are you basing your question on $$\sqrt[6]{7\sqrt[3]{20} - 19} = \sqrt[3]{\frac{5}{3}} - \sqrt[3]{\frac{2}{3}}$$ and the similar identities in this post? – Tito Piezas III Apr 14 '15 at 03:07
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@Tito Piezas III: Thank you for the reference. I have not known those. I don't immediately see how to answer my question, using the methods from the references in this post. – user64494 Apr 14 '15 at 04:09
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After finding my answer, I think there is a connection between, $$\sqrt[3] {-1+7\sqrt[3]{20}} =\sqrt[3]{\frac{16}{9}}+\sqrt[3]{\frac{100}{9}}-\sqrt[3]{\frac{5}{9}}\tag1$$ and $$\sqrt[3] {-19+7\sqrt[3]{20}} =\sqrt[3]{\frac{4}{9}}+\sqrt[3]{\frac{25}{9}}-\sqrt[3]{\frac{80}{9}}\tag2$$ Where did you find the LHS of $(1)$? – Tito Piezas III Apr 14 '15 at 05:45
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@Tito Piezas III: Share you opinion. The question originates from math folklore. – user64494 Apr 14 '15 at 06:49
1 Answers
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Yes.
$$ \sqrt[3] {-1+7\sqrt[3]{20}} =\sqrt[3]{\frac{16}{9}}+\sqrt[3]{\frac{100}{9}}-\sqrt[3]{\frac{5}{9}}\tag0$$
Solution: More generally, given the three roots $x_i$ of any cubic equation,
$$x^3+ax^2+bx+c=0\tag1$$
then sums involving the cube roots of the $x_i$ can be given in the simple form,
$$(u+x_1)^{1/3}+(u+x_2)^{1/3}+(u+x_3)^{1/3} = \big(w+3\,\sqrt[\color{blue}6]{d}\big)^{1/3}$$
where $u,w$ are the constants,
$$u = \frac{ab-9c+\sqrt{d}}{2(a^2-3b)}\tag2$$
$$w = -\frac{(2a^3-9ab+27c)+9\sqrt{d}}{2(a^2-3b)}\tag3$$
and $d$ is,
$$d = \tfrac{1}{27}\Bigl(4(a^2-3b)^3-(2a^3-9ab+27c)^2\Bigr)\tag4$$
Example: For your question, we have,
$$w=-1, \quad d =\frac{7^6\cdot20^2}{3^6}$$
and subbing these into $(3),(4)$, and using $Mathematica$ to simplify, we get $b,c$ for arbitrary $a$ as,
$$b= \tfrac{1}{9}(-343+3a^2)$$
$$c= \tfrac{1}{27}(-2058-343a+a^3)$$
Substituting $b,c,d$ into $(2)$ and $(1)$, we find that,
$$u=\tfrac{1}{9}(37+3a)$$
and $(1)$ factors as,
$$ (7 + a + 3 x) (14 + a + 3 x) (-21 + a + 3 x) = 0\tag5$$
giving $x_1, x_2, x_3$. The expression $u+x_1$ simplifies as just $\frac{16}{9}$ and similarly for $x_2, x_3$, thus resolving into the numerical relation $(0)$ given above.
Tito Piezas III
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thats a nice identity involving cube roots of roots of a cubic equation. +1 – Paramanand Singh Apr 14 '15 at 16:25
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@ParamanandSingh: Going to fifth powers, I have tried and have not found any example of $\sum_{n=1}^5 (u+x_n)^{1/5} = (w+\sqrt[k]{d})^{1/5}$ where the $x_i$ are five irrational roots of a quintic. – Tito Piezas III Apr 15 '15 at 02:33
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Sorry, but I am still not understanding what you mean by arbitrary $a$. Do you actually find the value of $a$? Or is it just a random number? (Both are definitions of arbitrary) – Frank May 31 '16 at 16:15
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@Frank. No. Give me a few minutes, and I'll add an example to the post. – Tito Piezas III May 31 '16 at 16:45
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Okay, tell me when it's finished... $\sqrt[3]{2}+\sqrt[3]{9}-\sqrt[3]{12}=\sqrt[3]{21\sqrt[3]{6}-37}$ – Frank May 31 '16 at 16:57
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@Frank: Yes, you are doing something wrong. Note that $d$ is a square so you should not get $\sqrt{105}$. Anyway, I have given enough help already so I hope you'll be able to answer further questions on your own. – Tito Piezas III Jun 01 '16 at 00:56
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@TitoPiezasIII Another one similar(?) to $\sqrt[6]{7\sqrt[3]{20}-19}=\sqrt[3]{\frac 53}-\sqrt[3]{\frac 23}$:$$\sqrt[6]{3\left(\frac {2-\sqrt[3]{4}}{2+\sqrt[3]{4}}\right)}=\sqrt[3]{\frac 43}-\sqrt[3]{\frac 13}$$ – Frank Nov 04 '16 at 00:12