Question

How do you factor `a^6+1`?

Answer 1

Factor as a sum of cubes. Then we can factor the remaining quartic factor into two quadratics:

`a^6+1`

`= (a^2 + 1)(a^4-a^2+1)`

`=(a^2+1)(a^2-sqrt(3)a+1)(a^2+sqrt(3)a+1)`

Explanation:

The sum of cubes identity can be written:

`A^3+B^3 = (A+B)(A^2-AB+B^2)`

So (putting `A=a^2` and `B=1`) we find:

`a^6+1 = (a^2)^3 + 1^3 = (a^2+1)((a^2)^2-a^2+1)`

`= (a^2 + 1)(a^4-a^2+1)`

It is not possible to factor `(a^2+1)` into linear factors with Real coefficients, so let's leave that alone and look at the remaining quartic factor `(a^4-a^2+1)`. This does not factor as a quadratic in `a^2` with Real coefficients, but will factor into two quadratics with irrational coefficients:

In order to get the terms in `a^4`, `a^3`, `a` and the constant term to work out, the factorisation must be something like this:

`a^4-a^2+1 = (a^2-ka+1)(a^2+ka+1)`

Then the coefficient of `a^2` is `2-k^2 = -1`. Hence `k = +-sqrt(3)`.

`a^4-a^2+1 = (a^2-sqrt(3)a+1)(a^2+sqrt(3)a+1)`

Both of these quadratic factors has negative discriminants, so only Complex roots.

Answer 2

Alternatively, use Complex arithmetic to find the linear factors, then combine in conjugate pairs to derive the Real factoring:

`a^6+1 = (a^2+1)(a^2-sqrt(3)+1)(a^2+sqrt(3)+1)`

Explanation:

Using De Moivre's Theorem:

`(cos theta + i sin theta)^n = cos (n theta) + i sin (n theta)`

Hence, the following Complex numbers are all `6`th roots of `-1`:

`cos(pi/6) + i sin(pi/6) = sqrt(3)/2 + 1/2 i`

`cos(pi/2) + i sin(pi/2) = i`

`cos((5pi)/6) + i sin((5pi)/6) = -sqrt(3)/2 + 1/2 i`

`cos((7pi)/6) + i sin((7pi)/6) = -sqrt(3)/2 - 1/2 i`

`cos((3pi)/2) + i sin((3pi)/2) = -i`

`cos((11pi)/2) + i sin((11pi)/2) = sqrt(3)/2 - 1/2 i`

graph{(x^2+(y-1)^2-0.0015)(x^2+(y+1)^2-0.0015)((x-sqrt(3)/2)^2+(y-1/2)^2-0.0015)((x-sqrt(3)/2)^2+(y+1/2)^2-0.0015)((x+sqrt(3)/2)^2+(y+1/2)^2-0.0015)((x+sqrt(3)/2)^2+(y-1/2)^2-0.0015) = 0 [-2.5, 2.5, -1.25, 1.25]}

Taking these in conjugate pairs we find quadratic factors with Real coefficients:

`(a - i)(a+i) = a^2 + 1`

`(a - (sqrt(3)/2 + 1/2 i))(a - (sqrt(3)/2 - 1/2 i)) = a^2 - sqrt(3)a + 1`

`(a + (sqrt(3)/2 + 1/2 i))(a + (sqrt(3)/2 - 1/2 i)) = a^2 + sqrt(3)a + 1`

Hence the factorisation with Real Coefficients is:

`a^6+1 = (a^2+1)(a^2-sqrt(3)+1)(a^2+sqrt(3)+1)`