# Taking the cube root of a sum of radicals

I am wondering how to derive the following simplification without knowing it beforehand: $$^3\sqrt{10 + 6\sqrt{3}} = 1 + \sqrt{3}$$ After the fact, it is easy to verify algebraically. The problem arose when applying Cardano's method to solve $$y^3 + 6y = 20$$ I was able to derive a similar but less complicated simplification, $$\sqrt{3 + 2\sqrt{2}} = 1 + \sqrt{2}$$ by assuming that $3 + 2\sqrt{2} = (a + b\sqrt{2})^2$, expanding, equating coefficients of $\sqrt{2}$, and solving for $a$ and $b$ using the quadratic formula. However, using the same method, i.e. assuming that $10 + 6\sqrt{3} = (a + b\sqrt{3})^3$, expanding, and equating coefficients of $\sqrt{3}$ yields cubic equations in $a$ and $b$: $$a^3 + 9ab^2 = 10, a^2b + b^3 = 2$$ Guess-and-check yields $a = 1$ and $b = 1$, but I would prefer a more systematic method of solution. I did not use the cubic formula again, figuring that this would probably yield another nested radical.

According to Wolfram|Alpha, $10 + 6\sqrt{3} = 1 + 3\sqrt{3} + 9 + 3\sqrt{3} = 1 + 3\sqrt{3} + 3(\sqrt{3})^2 + (\sqrt{3})^3 = (1 + \sqrt{3})^3$. However, I'm not sure how I would arrive at that chain of reasoning except by chance or by using Wolfram|Alpha.

• Your method gives $\{a^3+9ab^2=10, a^2b+b^3=2\}$. At the very least you could try the integer values by hand; the first equation gives $a$ as a factor of $10$, the second gives $b$ as a factor of $2$. – vadim123 Mar 7 '15 at 1:08
• @vadim123 Well, sure, I know that I could solve it by guess-and-check, but I was wondering if there was a more systematic method. – Radon Rosborough Mar 7 '15 at 1:12
• If $a,b$ aren't integers, you wouldn't be too pleased with the result anyway. – vadim123 Mar 7 '15 at 1:17
• @vadim123 Well, sure; I would preferably be looking for a method that would yield $a$ and $b$ always; then, when they are not integers I would know that the expression could not be simplified in this manner. – Radon Rosborough Mar 7 '15 at 1:19

$N(10+6\sqrt{3})=10^2-3\cdot6^2=-8$. So in $\mathbb Z[\sqrt3]$, it is possible that $10+6\sqrt{3}$ is a perfect cube. The only possible solution to $\sqrt{10+6\sqrt{3}}$ will be one with norm $-2$. That yields possible cube roots $(1\pm \sqrt{3})(2\pm \sqrt{3})^k$ for some $k$.
• Okay, so after some substantial research, I understand what a quadratic field is; where the formula for the norm, $N(x + y\sqrt{d}) = x^2 - dy^2$, comes from; and that the norm is multiplicative and so any possible cube root $a$ must have $N(a) = -2$. Also, I recognize that $(2 \pm \sqrt{3})^k$ are the units of $\mathbb{Z}[\sqrt{3}]$. But how do you find $1 \pm \sqrt{3}$ as a quadratic integer with norm $-2$ without solving the Diophantine equation $x^2 - 3y^2 = -2$, which would in general be difficult? – Radon Rosborough Mar 12 '15 at 23:47