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Using the Cardano formula, one can show that $\sqrt[3]{20+14\sqrt{2}}+\sqrt[3]{20-14\sqrt{2}}$ is a real root of the depressed cubic $f(x)=x^3-6x-40$. Actually, one can show by the calculating the determinant that this is the only real root. On the other hand, by the rational root theorem, one can see that the possible rational root must be a factor of 40 and one can check that $f(4)=0$. Therefore, by uniqueness of the real root, one must have $$ \sqrt[3]{20+14\sqrt{2}}+\sqrt[3]{20-14\sqrt{2}}=4\tag{1} $$

I had the observation above when I solved the cubic equation $x^3-6x-40=0$. My question is as follows:

without referring to the unique real root of the cubic, can we show (1) directly?


[An attempt.] When taking the cube on both sides of (1) and simplifying further, I ended up with (1) again.

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    $\begingroup$ This is almost a duplicate. See user8277998 answer to this question. $\endgroup$ – Dietrich Burde Aug 13 '18 at 18:10
  • $\begingroup$ One way is to denest both radicals and show that the irrational parts cancel out. $\endgroup$ – Frank W. Aug 13 '18 at 18:12
  • $\begingroup$ or this one math.stackexchange.com/questions/374619/… $\endgroup$ – mercio Aug 13 '18 at 18:12
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    $\begingroup$ without referring to the unique real root of the cubic The easiest way is in fact to derive the cubic from the given radicals (you don't have to know the cubic in advance). $\endgroup$ – dxiv Aug 13 '18 at 19:13
  • $\begingroup$ @dxiv: considering the answers below and the related links given in the comments above, I agree with you very much: solving a system of polynomial equations could be rather non-trivial, though in this particular case, a simple observation seems enough. The accepted answer in the link given by Dietrich seems mysterious to me: I don't see how one can "Solving the system" easily. $\endgroup$ – user486939 Aug 14 '18 at 15:07
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Well, $(2\pm\sqrt2)^3=20\pm14\sqrt2$, that is $\sqrt[3]{20\pm14\sqrt2} =2\pm\sqrt2$. Therefore $$\sqrt[3]{2+14\sqrt2}+\sqrt[3]{20-14\sqrt2}=2+\sqrt2+2-\sqrt2=4.$$

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  • $\begingroup$ Thanks for your answer! Would you explain how you came up with the nice identity $(2\pm\sqrt{2})^3=20\pm14\sqrt{2}$? I can check it directly though, I'm curious about what is the underlying motivation. $\endgroup$ – user486939 Aug 13 '18 at 23:35
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Can we express $\sqrt[3]{20+14\sqrt2}$ as $a+b\sqrt2$, with $a,b\in\mathbb Z$? In other words, are thre integers $a$ and $b$ such that $(a+b\sqrt2)^3=20+14\sqrt2$? Note that\begin{align}(a+b\sqrt2)^3=20+14\sqrt2&\iff a^3+3\sqrt2a^2b+6ab^2+2\sqrt2b^3=20+14\sqrt2\\&\iff a^3+6ab^2+(3a^2b+2b^3)\sqrt2=20+14\sqrt2.\end{align}Therefore, it is enough to have$$a^3+6ab^2=20(\iff a(a^2+6b^2)=20)\text{ and }3a^2b+2b^3=14(\iff b(3a^2+2b^2)=14).$$It is easy to see that you can take $a=2$ and $b=1$. Therefore, $\sqrt[3]{20+14\sqrt2}=2+\sqrt2$ and it is now easy to see that $\sqrt[3]{20-14\sqrt2}=2-\sqrt2$. Therefore$$\sqrt[3]{20+14\sqrt2}+\sqrt[3]{20-14\sqrt2}=2+\sqrt2+2-\sqrt2=4.$$

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    $\begingroup$ I like that you have provided a natural construction rather than just giving an equality. Thanks! $\endgroup$ – Clayton Aug 13 '18 at 18:36
  • $\begingroup$ Thanks for your answer! I was wondering how you observe that $\sqrt[3]{20+14\sqrt{2}}$ may be expressed as $a+b\sqrt{2}$. (I guess if $x^3=20+14\sqrt{2}$, then one may expect that $x$ is a $\mathbb{Z}$-linear combination of $1$ and $\sqrt{2}$?) $\endgroup$ – user486939 Aug 14 '18 at 0:54
  • $\begingroup$ @Mars Since the original equation has an integer root, I thought that it would be reasonable to expect that $20\pm14\sqrt2$ hade a cube root of the form $a\pm b\sqrt2$, with $a,b\in\mathbb Z$. $\endgroup$ – José Carlos Santos Aug 14 '18 at 6:31
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If you don't spot the values of these cube roots and don't feel like computing them, you can try this. Define $a_\pm:=(20\pm14\sqrt{2})^{1/3},\,s:=\sum_\pm a_\pm$ so $\sum_\pm a_\pm^3=40$ and $\prod_\pm a_\pm=8^{1/3}=2$, so $s^2=\frac{40}{s}+4ab=\frac{40}{s}+6$. Hence $0=s^3-6s-40=(s-4)(s^2+4s+10)$ has only one real root, $4$.

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  • $\begingroup$ Thanks for your answer. I think your factorization in the "Hence" step depends on knowing in advance that $4$ is a root of the polynomial? $\endgroup$ – user486939 Aug 14 '18 at 15:01
  • $\begingroup$ @Mars You can solve a cubic by Cardano's method, but we all prefer to spot a factorisation, preferably with rational coefficients. Not only is it worth a try; when you add a radical to its conjugate, you'd expect a rational result. So naturally, you use the rational root theorem to find candidates. $\endgroup$ – J.G. Aug 14 '18 at 15:11
  • $\begingroup$ I wrote in my post that "On the other hand, by the rational root theorem, one can see that the possible rational root must be a factor of 40 and one can check that f(4)=0." and I'm looking for an alternative proof for the identity. Anyway, thank you for your answer and comment again ;-) $\endgroup$ – user486939 Aug 15 '18 at 12:15

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