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Find the roots of $(z-1)^6 +(z+1)^6$. So far we've tried binomial expansion, but where to go now, as it is a non-calculator question?

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Hint: Consider the equation $\left(\frac{z-1}{z+1}\right)^6=-1$.

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This is equivalent to finding the roots of $$2(z^6 + 15z^4 + 15z^2 + 1) = 0.$$ If we substitute $x = z^2$, we get $x^3 + 15x^2 + 15x + 1 = 0$, which can be factored as $(x + 1)(x^2 + 14x + 1) = 0$. This can now be solved, and now we can use our substitution $x = z^2$ to find the roots of $z$ from the roots of $x$.

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I fixed some typo's. – Claude Leibovici Mar 10 '14 at 5:51
@ClaudeLeibovici Thank you. :) – laughing_man Mar 10 '14 at 18:54
You are welcome. – Claude Leibovici Mar 10 '14 at 19:16

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