Find the roots of a^6+a^4+a^2+1=0?

Question

How the find all the 6 roots of the above equation...??? please note that all the solutions are in-form of complex numbers.... i have the answers, but don't know how to arrive at 6 solutions to the above equation

Answer 1

a^6 + a^4 + a^2 + 1 = 0

a^4(a^2 + 1) + a^2 + 1 = 0

(a^2 + 1)(a^4 + 1) = 0

I'll leave the rest up to you.

EDIT: Time's up! Did you get

a^2 = -1 implies a = ± √(-1) = ± i

and

a^4 = -1 implies a^2 = ± i , implies a = ± √(± i)

Answer 2

a^4(a²+1)+(a²+1) = 0

(a²+1)(a^4+1) = 0

(a²-i²)(a^4- i²) = 0

(a²-i²)(a²- i)(a² +i) = 0

etc

Answer 3

a^6 + a^4 + a^2 + 1 = 0

(a^6 + a^4) + (a^2 + 1) = 0

a^4(a^2 + 1) + 1(a^2 + 1) = 0

(a^2 + 1)(a^4 + 1) = 0

a^2 + 1 = 0

a^2 = -1

a = ±√-1 (imaginary number)

(no real roots)

a^4 + 1 = 0

a^4 = -1

a = ±⁴√-1 (imaginary number)

(no real roots)

Answer 4

a^6 + a^4 + a^2 + 1 = 0

a^4 (a^2 + 1) + (a^2 + 1) = 0

(a^4 + 1)(a^2 + 1) = 0

(a^4 - i^2)(a^2 - i^2) = 0

(a^2 - i)(a^2 + i)(a - i)(a + i) = 0

a = +- i, +- √i