How to factor #a^8+b^8# ?
3 Answers
Explanation:
By the fundamental theorem of algebra, we can factor the given expression as
where each
Solving for
#=|b|(-1)^(1/8)# (assuming#b in RR# )
#=|b|(e^(i(pi+2pik)))^(1/8)#
#=|b|e^(ipi((2k+1)/8), k in ZZ#
As
For a more general
meaning
Sorry, I overlook some minor details, the answer provided by sente is correct.
Supposing
Define
and then
so