What is the value of $\sqrt[3]{9+2\sqrt5+3\sqrt{15}}$?

Question

$$\sqrt[3]{9+2\sqrt5+3\sqrt{15}}=?$$

Here is my work:

I considered the value is equal to $a+b$. Hence

$$9+2\sqrt5+3\sqrt{15}=(a+b)^3=a^3+b^3+3a^2b+3ab^2$$

$$2\sqrt5+3(3+\sqrt{15})=a^3+b^3+3(a^2b+ab^2)$$ We have $$ \begin{cases} {a^2b+ab^2=3+\sqrt{15}} \\ {a^3+b^3=2\sqrt5} \end{cases} $$

But I cannot find such $a$ and $b$.

Answer 1

I've got it, in the same sense as catching a tiger by the seat of my pants, but still ...

$\color{blue}{\sqrt[3]{9+2\sqrt5+3\sqrt{15}}=(1/\sqrt[3]2)(\frac32-\frac12\sqrt3+\frac12\sqrt5+\frac12\sqrt{15}).}$

This was checked numerically against the result claimed from WA.

Begin with the fact that the given radical actually has no solution of the form

$a+b\sqrt3+c\sqrt5+d\sqrt{15}$

with $a,b,c,d$ rational. In group theory language, we are denying a solution in $\mathbb Q[\sqrt3,\sqrt5]$. For suppose we did have

$\sqrt[3]{9+2\sqrt5+3\sqrt{15}}=a+b\sqrt3+c\sqrt5+d\sqrt{15}$

for some rational $a,b,c,d$. Then we would also have the conjugated results

$\sqrt[3]{9-2\sqrt5-3\sqrt{15}}=a+b\sqrt3-c\sqrt5-d\sqrt{15}$

$\sqrt[3]{9+2\sqrt5-3\sqrt{15}}=a-b\sqrt3+c\sqrt5-d\sqrt{15}$

$\sqrt[3]{9-2\sqrt5+3\sqrt{15}}=a-b\sqrt3-c\sqrt5+d\sqrt{15}$

Multiplying the right sides of these four equations would cancel the conjugated square roots so the product would be a rational number, therefore the product of the conjugated radicands on the left should be a rational cube. Unfortunately, we have instead

$(9+2\sqrt5+3\sqrt{15})(9-2\sqrt5-3\sqrt{15})(9+2\sqrt5-3\sqrt{15})(9-2\sqrt5+3\sqrt{15})=-5324=4×(-11)^3$

which, because of the factor of $4$ which is not a cube, fails to be a rational cube.

But, there is a way around this block. Suppose we were to double the radicand, to give $18+4\sqrt5+6\sqrt{15}$. To balance we attach a factor of $1/\sqrt[3]2$ outside the nested cube root. Now the product of the four conjugated radicands is $16$ times as large as before, which incorporates that factor of $4$ into the cube:

$(18+4\sqrt5+6\sqrt{15})(18-4\sqrt5-6\sqrt{15})(18+4\sqrt5-6\sqrt{15})(18-4\sqrt5+6\sqrt{15})=-85184=(-44)^3$

And we suppose a form

$\sqrt[3]{9+2\sqrt5+3\sqrt{15}}=(1/\sqrt[3]2)(\sqrt[3]{18+4\sqrt5+6\sqrt{15}})=(1/\sqrt[3]2)(a+b\sqrt3+c\sqrt5+d\sqrt{15})$

with $a,b,c,d$ rational.

We can do slightly better than this. The radicand is an algebraic integer, so the cube root must also be one. Thus, if the modified nested radical is indeed in $\mathbb Q[\sqrt3,\sqrt5]$, it must be in the algebraic integers of this group, thus $\mathbb Z[\sqrt3,\phi]$ where $\phi=(1+\sqrt5)/2$. So we seek

$\sqrt[3]{18+4\sqrt5+6\sqrt{15}}=k+l\sqrt3+m\phi+n\phi\sqrt3=(k+\frac12m)+(l+\frac12m)\sqrt3+\frac12m\sqrt5+\frac12n\sqrt{15}$

where $k,l,m,n$ are integers.

By trying integers with small absolute values and pure dumb luck, I hit on $k=1,l=-1,m=1,n=1$. So

$\sqrt[3]{18+4\sqrt5+6\sqrt{15}}=1-\sqrt3+\phi+\phi\sqrt3=\frac32-\frac12\sqrt3+\frac12\sqrt5+\frac12\sqrt{15}$

And therefore, with the balancing factor $1/\sqrt[3]2$ included, I get the result in the Spoiler.

Answer 2

Remark: Write $$9+2\sqrt{5}+3\sqrt{15}=(a+b\sqrt{3}+c\sqrt{5})^3.$$ Then the equations in rational $a,b,c$ yield immediately a contradiction.

Of course, there might be more complicated expressions. But for such questions, usually the answer is simple.

Edit: The equations are

\begin{align*} 0 & =6abc - 3\\ 0 & = 3a^2c + 9b^2c + 5c^3 - 2\\ 0 & =3a^2b + 3b^3 + 15bc^2\\ 0 & =a^3 + 9ab^2 + 15ac^2 - 9. \end{align*}