- Thread starter
- #1

\(\displaystyle x^8+1= P_0^3(x^2-2xcos\frac{(2k+1)\pi}{8}+1)\) where \(\displaystyle P_0^3\) means product from \(\displaystyle k=0\) to \(\displaystyle k=3\).

I tried it but got \(\displaystyle x^8+1= P_0^3(x^2-a_k^2)\) where \(\displaystyle a_k=cos\frac{(2k+1)\pi}{8}+isin\frac{(2k+1)\pi}{8}\). How to arrive at the correct answer