Value of an expression with cube root radical

Question

What is the value of the following expression?

$$\sqrt[3]{\ 17\sqrt{5}+38} - \sqrt[3]{17\sqrt{5}-38}$$

Answer 1

It is $2\sqrt{5}$. Note that $(17\sqrt{5}+38)^{\frac{1}{3}}=2+\sqrt{5}$.

Answer 2

Let $u=\sqrt[3]{38+17\sqrt{5}}$, $v= \sqrt[3]{38-17\sqrt{5}}$

From the equation $$u+v=n$$ and cubing, we obtain $$u^3+v^3+3uv(u+v)=n^3$$ that is $$76-3n=n^3$$ A root of this equation is $4$. In fact, it has no more real roots.

So we know that $u+v=4$, but we need $u-v$. We also know that $uv=-1$. Then $$u-v=\sqrt{(u+v)^2-4uv}=\sqrt{16+4}=2\sqrt 5$$

Answer 3

Here's the thing about this question: It is a trick.For most numbers of the form $a+b\sqrt 5$ the cube root is a mess. But some such numbers have nice cube roots and when they do we can find them.

The one thing you have to know is that if $x$ is any number of the form $a+b\sqrt5$ then $x^2, x^3$, etc. also have that form. For example: $$\begin{align} (a+b\sqrt 5)^2 & = a^2 + 2ab\sqrt 5 + 5b^2 \\ & = (a^2 + 5b^2) + 2ab\sqrt 5 \end{align}$$ Or similarly $$(a+b\sqrt 5)^3 = (a^3 + 15ab^2) + (3a^2b+5b^3)\sqrt 5.\tag{$\star$}$$

If $17\sqrt5 + 38$ has a simple cube root, it will be something of the form $a+b\sqrt 5$. From $(\star)$ we would want $$\begin{align} a^3+ 15ab^2 & = 38 \\ 3a^2b+5b^3 &= 17 \\ \end{align}$$

Looking at the first equation we see that maybe $15ab^2 = 30$ is almost 38, and then $a=2, b=1$ would work. This also works in the second equation. (Or we could look at the second equation first and see right away that the only way to get $3x+5y=17$ is to have $x=4, y=1$, so $a=2, b=1$.) This is called solving equations "by inspection". It is not quite a lucky guess; it is a thoughtful lucky guess based on the hope that the solution will be simple. But it also wouldn't be hard to try every possibility for $a$ and $b$: they can't be too big because then $a^3+15ab^2$ would be much bigger than 38.

Now we have $a=2, b=1$ we know that $\sqrt[3]{17\sqrt 5+ 38} = 2+\sqrt 5$. We can use the same method to find $\sqrt[3]{17\sqrt 5- 38}$ exactly, and from there the solution is easy.

Answer 4

Start by noticing that $(2 + \sqrt{5})^{3} = 38 + 17 \sqrt{5}$ and $(2 - \sqrt{5})^{3} = 38 - 17 \sqrt{5}$. Now, \begin{align} \sqrt[3]{\ 17\sqrt{5}+38} - \sqrt[3]{17\sqrt{5}-38} &= \sqrt[3]{(2 + \sqrt{5})^{3}} - \sqrt[3]{(\sqrt{5} - 2)^{3}} \\ &= (2 + \sqrt{5}) - (-2 + \sqrt{5}) \\ &= 4. \end{align}

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What seems to have happened is that most answers presented here follow the pattern: \begin{align} \sqrt[3]{\ 17\sqrt{5}+38} + \sqrt[3]{17\sqrt{5}-38} &= \sqrt[3]{(2 + \sqrt{5})^{3}} + \sqrt[3]{(\sqrt{5} - 2)^{3}} \\ &= (2 + \sqrt{5}) + (-2 + \sqrt{5}) \\ &= 2 \sqrt{5}. \end{align}

Answer 5

$$\sqrt[3]{\ 17\sqrt{5}+38} - \sqrt[3]{17\sqrt{5}-38}=$$ $$\sqrt[3]{8+12\sqrt{5}+30+5\sqrt{5}} - \sqrt[3]{-8+12\sqrt{5}-30+5\sqrt{5}}=$$ $$\sqrt[3]{8+12\sqrt{5}+6\left(\sqrt{5}\right)^2+\left(\sqrt{5}\right)^3} - \sqrt[3]{-8+12\sqrt{5}-6\left(\sqrt{5}\right)^2+\left(\sqrt{5}\right)^3}=$$ $$\sqrt[3]{\left(\sqrt{5}+2\right)^3} - \sqrt[3]{\left(\sqrt{5}-2\right)^3}=$$ $$\left({\left(\sqrt{5}+2\right)^3}\right)^{\frac{1}{3}} - \left({\left(\sqrt{5}-2\right)^3}\right)^{\frac{1}{3}}=$$ $$\left(\sqrt{5}+2\right) -\left(\sqrt{5}-2\right)=$$ $$\left(\sqrt{5}+2\right) +\left(2-\sqrt{5}\right)=$$ $$2+2+\sqrt{5}-\sqrt{5}=2+2=4$$

Answer 6

The expression strangely reminds of Cardano's formula for the cubic equation

$$x=\sqrt[3]{-\frac q2-\sqrt{\frac{q^2}4+\frac{p^3}{27}}}+\sqrt[3]{-\frac q2+\sqrt{\frac{q^2}4+\frac{p^3}{27}}}.$$

Identifying, we have

$$q=-76,p=3,$$

and $x$ is a root of $$x^3+3x-76=0.$$ By inspection (trying values close to $\sqrt[3]{76}$), $\color{green}4$ is a solution.

As $$x^3+3x-76=(x-4)(x^2+4x+19),$$ the other roots are complex.

Answer 7

\begin{align} x&=\sqrt[3]{17\sqrt{5}+38} - \sqrt[3]{17\sqrt{5}-38} \end{align}

Note that \begin{align} 17\sqrt{5}-38&=\frac{1}{17\sqrt{5}+38}. \end{align}

Let $a=\sqrt[3]{17\sqrt{5}+38}$. Then we have \begin{align} x^3&=\left(a-\frac1a\right)^3 \\ x^3&= a^3-3a+\frac3a-\frac{1}{a^3} \\ x^3+3\left(a-\frac1a\right)&= a^3-\frac{1}{a^3} \\ x^3+3x-76&=0 \\ (x-4)(x^2+4x+19)=0, \end{align} The only real solution $x=4$ is the answer.