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I wonder whether there is a pattern that goes on and on: $$(a+b)\,(a-b) = a^2 - b^2$$ $$(a+b)\,(a+(-1/2 + i \sqrt{3}/2)b)\,(a+(-1/2 - i \sqrt{3}/2)b) = a^3 + b^3$$ $$(a+b)\,(a+i b)\,(a-b)\,(a-i b) = a^4 - b^4$$ The general product would be as follows where $\epsilon = e^{2 i \pi/n}$ is the n-th unit root: $$\prod_{k=0}^{n-1}(a+\epsilon^k b) =\,?$$

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    $\begingroup$ $a^n-(-b)^n\phantom{}$. $\endgroup$ – Lord Shark the Unknown Jun 23 '17 at 20:26
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    $\begingroup$ Any proof for that? It looks like, but I am not sure. $\endgroup$ – j4n bur53 Jun 23 '17 at 20:28
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    $\begingroup$ Use \prod for $\prod$. It works better than \Pi :) $\endgroup$ – Cameron Williams Jun 23 '17 at 20:37
  • $\begingroup$ Hint: $\prod_{k=0}^{n-1}(a+\epsilon^k b) = (-1)^nb^n \prod_{k=0}^{n-1}\left(\frac{a}{b}-\epsilon^k\right)=(-1)^nb^nP\left(\frac{a}{b}\right)$ where $P(z)$ is the polynomial with roots $1, \epsilon, \epsilon^2, \cdots, \epsilon^{n-1}\,$. $\endgroup$ – dxiv Jun 23 '17 at 20:37
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Your problem is equivalent to find the roots of

$$p(a)=a^n-(-1)^nb^n$$

consider $b>0$. So, the roots are $$-b\cdot (\text {roots of unit})$$

and once you can split $p(a)$ as:

$$(a-a_1)(a-a_2)...(a-a_n)$$

where $a_i$ is a root of $p(a)$ then you get what you want.

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