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.

  • $\begingroup$ 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 $2$. $\endgroup$ – vadim123 Mar 7 '15 at 1:08
  • $\begingroup$ @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. $\endgroup$ – Radon Rosborough Mar 7 '15 at 1:12
  • $\begingroup$ If $a,b$ aren't integers, you wouldn't be too pleased with the result anyway. $\endgroup$ – vadim123 Mar 7 '15 at 1:17
  • $\begingroup$ @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. $\endgroup$ – 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[3]{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 $k$.

  • $\begingroup$ I don't think I have the requisite background in polynomial rings (?) to fully understand this solution. For instance, where does the formula for the norm come from? (Is there somewhere I can find this information online?) $\endgroup$ – Radon Rosborough Mar 7 '15 at 1:38
  • 1
    $\begingroup$ That's probably too big a topic to cover in an answer here, but the topics of "quadratic fields" and "quadratic integers" are the place to start. @raxod502 $\endgroup$ – Thomas Andrews Mar 7 '15 at 3:03
  • $\begingroup$ 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? $\endgroup$ – Radon Rosborough Mar 12 '15 at 23:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.