# Generalization of conjugate $a-b$ for $a+b$

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: $$\prod_{k=0}^{n-1}(a+\epsilon^k b) =\,?$$

• $a^n-(-b)^n\phantom{}$. – Lord Shark the Unknown Jun 23 '17 at 20:26
• Any proof for that? It looks like, but I am not sure. – j4n bur53 Jun 23 '17 at 20:28
• Use \prod for $\prod$. It works better than \Pi :) – Cameron Williams Jun 23 '17 at 20:37
• 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 $1, \epsilon, \epsilon^2, \cdots, \epsilon^{n-1}\,$. – dxiv Jun 23 '17 at 20:37

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