Question

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

Answer 1

`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)))`

Answer 2

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

Answer 3

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.