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How to show that the following is true? $$\sqrt[3]{26+15\sqrt{3}}+\sqrt[3]{26-15\sqrt{3}} \in \mathbb{Z}$$

I have tried to set $$\sqrt[3]{26+15\sqrt{3}}+\sqrt[3]{26-15\sqrt{3}} = r,$$ $$a=\sqrt[3]{26+15\sqrt{3}},$$ $$b=\sqrt[3]{26-15\sqrt{3}},$$ and used the identity $$(a^{1/3} + b^{1/3})^3 = a + b + 3(ab)^{1/3}(a^{1/3} + b^{1/3})$$ but I got nowhere. I am stuck at $$\left(a^{1/3}+b^{1/3}\right)^3=52+ 3 \cdot \left(a^{1/3}+b^{1/3}\right)$$ I'd be glad at any help.

user50224
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3 Answers3

4

We have $$ 26 \pm 15\sqrt{3} = (2 \pm \sqrt{3})^3 $$ (by educated guessing, see note below) which gives us $$ \sqrt[3]{26 + 15\sqrt{3}} + \sqrt[3]{26-15\sqrt3} = 2+\sqrt3+2-\sqrt3 = 4 $$

Note on guessing: I guessed that $26 + 15\sqrt3$ would be a nice cube of the form $(n + m\sqrt3)^3$ for integers (or at least rational numbers) $n, m$. If that were indeed the case, then we would have $$ 26 + 15\sqrt3 = (n + m\sqrt3)^3 = n^3 + 3n^2m\sqrt3 + 9nm^2 + 3m^3\sqrt3 $$ which then becomes $$ n^3 + 9nm^2 = 26 \qquad\bigwedge\qquad 3(n^2m + m^3) = 15 $$ The second equation becomes $(n^2 + m^2)m = 5$ which has $n = 2, m = 1$ as the nicest solution. It turns out that this solution also solves the first equation, so we're done.

Arthur
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  • This is exactly how I did it, so the “guessing” suggestion is not at all far-fetched. You might also observe that by changing $m, n$, and the radicand $3$ we can construct many other example of this type. For example taking $m=2, n=1$ and replacing $\sqrt3$ with $\sqrt 2$ we find that $$\sqrt[3]{25+22\sqrt2} + \sqrt[3]{25-22\sqrt2} = 2.$$ – MJD Jun 18 '23 at 02:49
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Set $ r=\sqrt[3]{26+15\sqrt{3}}+\sqrt[3]{26-15\sqrt{3}}$. Then $$r^3= 26+15\sqrt{3}+3\sqrt[3]{( 26+15\sqrt{3})^2}\sqrt[3]{ 26-15\sqrt{3}}+3\sqrt[3]{ 26+15\sqrt{3}}\sqrt[3]{( 26-15\sqrt{3})^2}+ 26-15\sqrt{3}=52+3( \sqrt[3]{26+15\sqrt{3}}+\sqrt[3]{ 26-15\sqrt{3}}\;)=52+3r.$$ Therefore $r^3-3r-52=0$. A quick factorization gives us $(r-4)(r^2+4r+13)=0$. Since $ r^2+4r+13$ doesn't have real roots, we deduce that $ r=4$.

Xam
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  • Thanks, but I still got one question: Until your factorization I understand it. But I am supposed to show that $r \in \mathbb{Z}$. So, how can I assume it to be a real number - which you do when you say that the quadratic equation does not have real roots... - Especially how do I know that $r$ is the real root of the cubic equation and not any other root? – user50224 Nov 05 '16 at 07:12
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    @user50224 $r$ is the sum of two cubic roots, hence $r$ is a real number. – Xam Nov 05 '16 at 15:16
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Find $p$ and $q$ such that $$ -\frac{q}{2}=26 \qquad \frac{q^2}{4}+\frac{p^3}{27}=675 $$ We get $q=-52$ and $p=-3$. So your number is a root of $$ x^3-3x-52=0 $$ and you can see that $4$ is the only real solution. Cardan's formula tells you that $$ \sqrt[3]{26+15\sqrt{3}}+\sqrt[3]{26-15\sqrt{3}}=4 $$

Alternatively, $$ \frac{1}{26-15\sqrt{3}}=\frac{26+15\sqrt{3}}{26^2-675}=26+15\sqrt{3} $$ so $b=a^{-1}$. Then $$ r^3=(a+a^{-1})^3=a^3+3a+3a^{-1}+a^{-3}=3r+52 $$

egreg
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