<s>[INST] What is $(-1)^{\frac{1}{7}} + (-1)^{\frac{3}{7}} + (-1)^{\frac{5}{7}} + (-1)^{\frac{9}{7}} + (-1)^{\frac{11}{7}}+ (-1)^{\frac{13}{7}}$? The question is as given in the title. According to WolframAlpha, the answer is 1, but I am curious as to how one gets that. I tried simplifying the above into $$6(-1)^{\frac{1}{7}}$$ however that does not give 1.
 [/INST] Note that $(-1)^{\frac{1}{7}} = -1$
Also note that $(-1)^{\frac{9}{7}} = -(-1)^{\frac{2}{7}},\ (-1)^{\frac{11}{7}} = -(-1)^{\frac{4}{7}},\ (-1)^{\frac{13}{7}} = -(-1)^{\frac{6}{7}}$
You can then rewrite your addition as
$$(-1)^{\frac{1}{7}} - (-1)^{\frac{2}{7}} + (-1)^{\frac{3}{7}} - (-1)^{\frac{4}{7}} + (-1)^{\frac{5}{7}} - (-1)^{\frac{6}{7}}$$
Now remembering that $a^{\frac{p}{q}} = (a^{\frac{1}{q}})^p = (\sqrt[q]{a})^p$ and substituting $q = 7$ and the appropriate $p$ you get:
$$(-1)^{\frac{1}{7}} - (-1)^{\frac{2}{7}} + (-1)^{\frac{3}{7}} - (-1)^{\frac{4}{7}} + (-1)^{\frac{5}{7}} - (-1)^{\frac{6}{7}} \\
= (-1) - (1) + (-1) - (1) + (-1) - (1) \\
= -6$$
Now what if instead of $a^{\frac{p}{q}} = (a^{\frac{1}{q}})^p$ you make $a^{\frac{p}{q}} = \sqrt[q]{a^p}$? Then for negative $a$, odd $p$ and even $q$ that would only be calculatable if you entered the realm of the complex numbers and made use of the imaginary unit. But when you enter the realm of the complex numbers, you get that $\sqrt[q]{x}$ has $q$ different values. That is, there are $q$ different numbers $x_i$ for which $(x_i)^q = x$. 
WolframAlpha goes straight ahead and uses the complex plane to calculate the roots making use of the complex numbers. Now if you make use of the fact that the sum of all $q$ roots of $x$ is 0, you can do the following:
Call $x_i$ the ith root of (-1). Then $x_i = (-1)^{\frac{i}{7}}$ and if you notice that in your sum all roots are represented except for $x_0$ which is -1, then
$$\sum_{i=0}^{6} x_i = -1 + \sum_{i=1}^{6} x_i$$ but $$\sum_{i=0}^{6} x_i = 0$$ so $$0 = -1 + \sum_{i=1}^{6} x_i \iff \sum_{i=1}^{6} x_i = 1$$
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