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

##### 2 Answers

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^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

In order to get the terms in

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

Then the coefficient of

#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.

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

#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)#