# Show that $\sqrt[3]{3\sqrt{21} + 8} - \sqrt[3]{3\sqrt{21} - 8} = 1$

Show that $$\sqrt[3]{3\sqrt{21} + 8} - \sqrt[3]{3\sqrt{21} - 8} = 1$$

Playing around with the expression, I found a proof which I will post as an answer.

I'm asking this question because I would like to see if there are alternative solutions which are perhaps faster / more direct / elementary / elegant / methodical / insightful etc.

$(\frac {1 \pm \sqrt{21}}2)^3 = 8 \pm 3\sqrt{21}$

These answers are also relevant I guess Is $\sqrt[3]{p+q\sqrt{3}}+\sqrt[3]{p-q\sqrt{3}}=n$, $(p,q,n)\in\mathbb{N} ^3$ solvable? How does one evaluate $\sqrt[3]{x + iy} + \sqrt[3]{x - iy}$?

• I accepted this solution because mercio was first to notice (in a comment at the question) that the cubic roots can indeed be simplified. – azimut Jan 9 '15 at 12:22
• +1. For the record: I hadn't seen mercio's comment before submitting my own solution. :) – Blue Jan 9 '15 at 12:26

Here is my solution:

Let $\alpha = \sqrt[3]{3\sqrt{21} + 8}$ and $\beta = \sqrt[3]{3\sqrt{21} - 8}$. Then $$\alpha\beta = \sqrt[3]{(3\sqrt{21} + 8)(3\sqrt{21} - 8)} = \sqrt[3]{(3\sqrt{21})^2 - 8^2} = \sqrt[3]{189 - 64} = \sqrt[3]{125} = 5$$ and $$\alpha^3 - \beta^3 = (3\sqrt{21} + 8) - (3\sqrt{21} - 8) = 16$$ Now $$(\alpha - \beta)^3 = \alpha^3 - 3\alpha^2\beta + 3\alpha\beta^2 - \beta^3 = (\alpha^3 - \beta^3) - 3\alpha\beta (\alpha - \beta) = 16 - 15(\alpha - \beta)$$ so $\alpha - \beta$ is a root of the polynomial $$x^3 + 15x - 16 = (x-1)(x^2 + x + 16).$$ The part $x^2 + x + 16$ does not have any real roots since its discriminant is $-1 - 4\cdot 16 < 0$. So $\alpha - \beta$ is a root of $x-1$ and thus $$\alpha - \beta = 1$$

• this is a nice one. but you want to prove $\alpha + \beta = 1,$ right? – Krish Jan 9 '15 at 11:29
• @Krish apparently the typo is in the OP: wolframalpha.com/input/… – Surb Jan 9 '15 at 11:32
• @Krish Sorry, there was a typo in the question. So it's really $\alpha - \beta = 1$. – azimut Jan 9 '15 at 11:57

Sometimes you can get lucky de-nesting a radical. This is one of those times.

Write $\sqrt[3]{3\sqrt{21}\pm 8}$ as $\frac12\;\sqrt[3]{24\sqrt{21}\pm 64}$, and consider expressing the radicand as a perfect cube: $$24\sqrt{21}\pm 64 = (\;p + q \sqrt{21}\;)^3 =p^3 + 3 p^2 q \sqrt{21} + 63 p q^2 + 21 q^3 \sqrt{21}$$ so that $$p\left(\; p^2 + 63 q^2 \;\right) = \pm 64 \qquad q \left(\;p^2 + 7 q^2\;\right)\cdot 3\sqrt{21} = 8 \cdot 3\sqrt{21}$$

Clearly, we can take $p = \pm 1$ and $q = 1$. Then, \begin{align} \sqrt[3]{3\sqrt{21}+8} - \sqrt[3]{3\sqrt{21}-8} &= \frac{1}{2}\left(\; \sqrt[3]{(1 + \sqrt{21} )^3} - \sqrt[3]{(-1 + \sqrt{21} )^3} \;\right) \\[6pt] &= \frac{1}{2}\left(\;1 + \sqrt{21} - (-1 + \sqrt{21})\;\right) \\[6pt] &= 1 \end{align}

As a bonus, we can also obtain that

$$T=\sqrt[3]{3\sqrt{21} + 8} + \sqrt[3]{3\sqrt{21} - 8}=\sqrt{21}.$$ Indeed, we can write $$T^3 = 6\sqrt{21}+6\sqrt{21}.$$ In order to get rid of $\sqrt{21}$ we can write $T = \alpha \sqrt{21}$ to get the equation $$21\alpha^3 - 15\alpha-6.$$ We have an obvious root $\alpha=1$, and then prove that $$\frac{21\alpha^3 - 15\alpha-6}{\alpha-1} = 21\alpha^2 +21\alpha+6$$does not have real roots.

• Thank you and (+1) for the bonus, which in some sense answers the first version of my question before I corrected the typo. – azimut Jan 9 '15 at 12:13

HINT: If we can recognize here Cardano's formula, finding the equation and its obvious root $x=1$ is easy.

let $\sqrt[3]{\sqrt{a}+b}-\sqrt[3]{\sqrt{a}-b}=1$ so that $a=189$ and $b=8$

$(\sqrt[3]{\sqrt{a}+b}-\sqrt[3]{\sqrt{a}-b})^3=1$

$\sqrt{a}+b-3(\sqrt[3]{(\sqrt{a}+b)^2}\sqrt[3]{(\sqrt{a}-b)})+3(\sqrt[3] {(\sqrt{a}+b)}\sqrt[3]{(\sqrt{a}-b)^2})-\sqrt{a}+b=1$

$2b-3(\sqrt[3]{(\sqrt{a}+b)^2}\sqrt[3]{(\sqrt{a}-b)})+3(\sqrt[3]{(\sqrt{a}+b)}\sqrt[3]{(\sqrt{a}-b)^2})=1$

$2b-3(\sqrt[3]{(\sqrt{a}+b)}\sqrt[3]{(\sqrt{a}-b)})[ \sqrt[3]{\sqrt{a}+b} -\sqrt[3]{\sqrt{a}-b}]=1$

$2b-3(\sqrt[3]{\sqrt{a}+b}.\sqrt[3]{\sqrt{a}-b})(1))=1$

$2b-3\sqrt[3]{a-b^2}=1$

$2*8-3\sqrt[3]{189-8^2}=1\rightarrow 16-15=1$