How to factor a^8+b^8 ?

3 Answers
Oct 10, 2016

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.