Find $\displaystyle\sqrt[3]{2+\frac {10} 9\sqrt 3}+\sqrt[3]{2-\frac {10} 9\sqrt 3}$.

I found that, by calculator, it is actually $\bf{2}$.

Methods to denest something like $\sqrt{a+b\sqrt c}$ seems to be useless here, what should I do?

  • $\begingroup$ Interesting. Looks like Cardan's formulas. $\endgroup$ – user135971 Apr 9 '15 at 10:10

We have

$\displaystyle 2 + \frac{10}{9} \sqrt{3} = 1 + \sqrt{3} + 1 + \frac{1}{9} \sqrt{3}$ $\displaystyle = 1 + \frac{3}{\sqrt{3}} + \frac{3}{\sqrt{3}^2} + \frac{1}{\sqrt{3}^3}= \bigl(1 + \frac{1}{\sqrt{3}}\bigr)^3$

Similarily, we have $2-\frac{10}{9} \sqrt{3} = \left(1-\frac{1}{\sqrt{3}}\right)^3$. Hence

$\displaystyle \sqrt[3]{2+\frac {10} 9\sqrt 3}+\sqrt[3]{2-\frac {10} 9\sqrt 3} ~ = ~ \bigl(1 + \frac{1}{\sqrt{3}}\bigr) +\bigl(1-\frac{1}{\sqrt{3}}\bigr) = 2$.

PS: I have to admit that the first line is a little bit unmotivated. You need some experience in order to see these transformations (whereas verification is trivial). Alternatively, you can hope that both summands of this complicated sum actually lie in $\mathbb{Q}(\sqrt{3})=\{p+q \sqrt{3} : p,q \in \mathbb{Q}\}$, make the Ansatz $2+\frac{10}{9} \sqrt{3} = (p+q \sqrt{3})^3=\dotsc$ and solve for $p,q$, which gives $p=1$ and $q=\frac{1}{3}$ as a possible solution. The answer by lab bhattacharjee explains how to calculate the sum more "mechanically".

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    $\begingroup$ How do you observe to get the result on first line? $\endgroup$ – JSCB May 9 '13 at 11:02
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    $\begingroup$ @Martin, not sure what do you mean by "mechanically"? Also, solving for $p,q$ is non-trivial, right? I think my solution is more generic & universal than this one? $\endgroup$ – lab bhattacharjee May 9 '13 at 16:05

$$S=\displaystyle\sqrt[3]{2+\frac {10} 9\sqrt 3}+\sqrt[3]{2-\frac {10} 9\sqrt 3}$$

So, $$S^3=2+\frac {10} 9\sqrt 3+2-\frac {10} 9\sqrt 3+3\cdot \sqrt[3]{2+\frac {10} 9\sqrt 3}\cdot\sqrt[3]{2-\frac {10} 9\sqrt 3}\cdot S$$

$$\implies S^3=4+2S$$ as $(2+\frac {10} 9\sqrt 3)(2-\frac {10} 9\sqrt 3)=4-\frac{100\cdot3}{81}=\frac8{27}=(\frac23)^3$

$$\iff S^3-2S-4=0$$

$$\implies S^3-2^3-2(S-2)=0\implies (S-2)(S^2+S+2)=0$$

Observe that $2$ is the sole positive real root of the last eqaution


Let $x=\sqrt[3]{2+\frac {10}{9}\sqrt 3}, y=\sqrt[3]{2-\frac {10}{9}\sqrt 3}$. Then $$ x^3+y^3=4,\quad xy=\frac{2}{3}. $$ The rest is yours.

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    $\begingroup$ Please explain how this can help the OP. $\endgroup$ – Did May 9 '13 at 10:17
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    $\begingroup$ @DId: I think putting $y=2x/3$ one would get the same cubic equation as lab bhattacherjee got. $\endgroup$ – pritam May 9 '13 at 10:21
  • $\begingroup$ @pritam You probably mean $y=2/(3x)$, not $y=2x/3$. Then $x^3+y^3=4$ becomes a quadratic (not cubic) equation in $x^3$ (not in $x$). Right, but what next? I recall that the object of interest here is $x+y$, neither $x$ nor $x^3$. $\endgroup$ – Did May 9 '13 at 10:46
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    $\begingroup$ But now it's easy: $(x+y)^3=x^3+y^3+3xy(x+y)$, so $(x+y)^3-2(x+y)-4=0$. Putting $S=x+y$, this is $(S-2)(S^2+2S+2)=0$, with one real solution $S=2$. $\endgroup$ – TonyK May 9 '13 at 11:45
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    $\begingroup$ Nice hint, the "Cardano" trick backwards. Maybe could have added "expand $(x+y)^3$." $\endgroup$ – André Nicolas May 9 '13 at 13:41

Find $p$ and $q$ such that

\begin{cases} -\frac{q}{2}=2\\[3ex] \frac{p^3}{27}+\frac{q^2}{4}=\frac{100}{27} \end{cases}

This of course gives $q=-4$ and

$$ \frac{p^3}{27}=\frac{100}{27}-4=-\frac{8}{27}, $$

so $p^3=-8$ and $p=-2$.

Now find the unique real root of the polynomial $x^3+px+q=x^3-2x-4$, which is of course $2$. Indeed the polynomial factor as

$$ x^3-2x-4=(x-2)(x^2+2x+2) $$ and the second degree factor has no real roots.

This example clearly shows why Cardan's formulas have a very limited usefulness: they present the roots in a form that makes them unintelligible.

Here's Cardan's formula for third degree equations with only one real root: if $x^3+px+q=0$ and the discriminant


then the only real solution for the equation is

$$ \sqrt[3]{-\frac{q}{2}+\sqrt{\frac{p^3}{27}+\frac{q^2}{4}}} + \sqrt[3]{-\frac{q}{2}-\sqrt{\frac{p^3}{27}+\frac{q^2}{4}}} $$

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    $\begingroup$ I would say that this example shows that one should not use Cardano's formula before testing for integer solutions. $\endgroup$ – Martin Brandenburg May 9 '13 at 10:40
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    $\begingroup$ @MartinBrandenburg In a sense this is how Tartaglia and Ferrari made their problems obscure: they prepared an equation from a given (non easy to spot) root. $\endgroup$ – egreg May 9 '13 at 10:44

Showing that $\displaystyle\sqrt[3]{2+\frac {10} 9\sqrt 3}+\sqrt[3]{2-\frac {10} 9\sqrt 3} = 2$ is the same as showing that $2\pm\frac {10} 9\sqrt 3$ has a cube root of the form $1 \pm a\sqrt 3$ (with rational $a$).

$(1\pm a\sqrt 3)^3 = (1+9a^2)\pm(3a+3a^3)\sqrt 3$, and you quickly check that with $a = 1/3$ you simultaneously get $1+9a^2 = 2$ and $3a(1+a^2) = \frac{10}9$, hence $2 \pm \frac{10}9\sqrt 3 = (1\pm\frac 13\sqrt 3)^3$, and your original expression simplifies to $1+\frac 13\sqrt 3+1-\frac 13\sqrt 3=2$


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