Show that $\sqrt[3]{2+\frac {10} 9\sqrt 3}+\sqrt[3]{2-\frac {10} 9\sqrt 3}=2$

Question

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?

Answer 1

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".

Answer 2

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.

Answer 3

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

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

Answer 4

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

$$\frac{p^3}{27}+\frac{q^2}{4}>0$$

then the only real solution for the equation is

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

Answer 5

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 $1+\frac 13\sqrt 3+1-\frac 13\sqrt 3=2$