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

##### 2 Answers

#### Answer:

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.

#### Answer:

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)#