How to factor #a^8+b^8# ?

3 Answers
Oct 10, 2016

Answer:

#a^8+b^8 = prod_(k=0)^7(a-|b|e^(ipi(2k+1)/8))# for #b in RR#

#a^8+b^8 = prod_(k=0)^7(a-|b|e^(ipi(theta/pi+(2k+1)/8)))# for #b = |b|e^(itheta) in CC#

Explanation:

By the fundamental theorem of algebra, we can factor the given expression as

#a^8+b^8 = prod_(k=1)^8(a-alpha_k)#

where each #alpha_k# is a root of #x^8+b^8#.

Solving for #alpha_k#, we get

#x^8+b^8 = 0#

#=> x^8 = -b^8#

#=> x = (-b^8)^(1/8)#

#=|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 #k in {0, 1, 2, 3, 4, 5, 6, 7}# accounts of all unique values of that form, we get our factorization as, for #b in RR#

#a^8+b^8 = prod_(k=0)^7(a-|b|e^(ipi(2k+1)/8))#


For a more general #b in CC#, then supposing #b = |b|e^(itheta)#, we can go through similar calculations to find

#(-b^8)^(1/8) = |b|e^(ipi(theta/pi+(2k+1)/8))#

meaning

#a^8+b^8 = prod_(k=0)^7(a-|b|e^(ipi(theta/pi+(2k+1)/8)))#

Oct 10, 2016

Sorry, I overlook some minor details, the answer provided by sente is correct.

Oct 11, 2016

Supposing #b ne 0# and #a,b in RR# we have
#(a/b)^8=-1 = e^(ipi+2kpi)# then
#a/b=e^(i(2k+1)pi/8)# then
#a-b e^(i(2k+1)pi/8)=0# are the #k=0,1,cdots,7# roots or factors.

Define

#p(k) = a-be^(i(2k+1)pi/8)#

and then

#f_1=p(1)p(6) =a^2 - (sqrt[2 - sqrt[2]]) a b + b^2#
#f_2=p(2)p(5)=a^2 + (sqrt[2 - sqrt[2]]) a b + b^2#
#f_3=p(3)p(4)=a^2 + (sqrt[2 + sqrt[2]]) a b + b^2#
#f_4 = p(0)p(7)=a^2 - (sqrt[2 + sqrt[2]]) a b + b^2#

so

#a^8+b^8=f_1 f_2 f_3 f_4# with real coefficients.