# Reducing a sum of two cube roots

I'm trying to verify that this sum of cube roots is equal to 1, using Solve, ToRoots, RootReduce, Expand, restricting to $$\mathbb{R}$$, and so on, but am failing:

$$\sqrt{8 + 3 \sqrt{21}} + \sqrt{8 - 3 \sqrt{21}}$$

Even N@... fails to give the answer.

I can solve this by doing lots of manipulation essentially "by hand," taking care with choice of branch cuts, and so on, but I would have thought Mathematica could reduce such an apparently simple sum of terms, which might involve the solution to a cubic, which of course the system "knows."

Suggestions?

• CubeRoot[8 + 3 Sqrt] + CubeRoot[8 - 3 Sqrt] // FullSimplify Sep 2 at 22:40
• Thanks ($+1$). But I still wonder why the straightforward method fails to find the right branch cut. Sep 2 at 22:45
• Factor out -1 from the second cube root, so MMA does not see a negative number raised to a fractional power, as ((8 + 3 Sqrt)^(1/3) - (-8 + 3 Sqrt)^(1/3)) // RootReduce Sep 2 at 22:55
• @LouisB: Good idea... but that still relies on "human smarts" which makes sense here with the odd-power root, but may not in other contexts. Sep 2 at 23:00
• The standard convention is to define $a^b = \exp(b\log(a))$. And so for example $(-1)^{1/3} = \exp(\log(-1)/3) = \exp(i\pi/3) = 1/2 + i\sqrt{3}/2$. Both CubeRoot and Surd use a non-principal branch of the logarithm. This reminds me of this video. youtube.com/watch?v=5pa1AryylpM Sep 3 at 15:12

Use Surd to get the branch you want.
Surd[(8 + 3 Sqrt), 3] + Surd[(8 - 3 Sqrt), 3] // RootReduce

• Oh... thanks for your quick answer ($\checkmark$). Do you have a quick explanation why using Surd works while the "traditional" way chooses the wrong branch cut? Sep 2 at 22:38