I've got this math problem in my head which I just can't seem to figure out.

Question:
"Find, in a + bi form, the two roots of the equation (1-x)^3 = x^3"

I've already found one root...

(1-x)^3 = x^3
(1-x) = x
1 = 2x
x = 1/2

But how would I go about getting the other root?

(1-x)^3 = x^3
1-x^3=x^3
1=x^3+x^3
1=x^6
1=x

If I'm remembering correctly

(1-x)^3 = 1-3x+3x^2-x^3

u trollin?

Anyway, Since the highest power of x is 3, there are 3 roots. As you found out, 1/2 is a root.

Expanding out the (1-x)^3 term and bringing everything to the same side, we get 2x^3-3x^2+3x-1 = 0
Factoring out (x-1/2), we get (x-1/2)*(2x^2-2x+2)=0, or equivalently (x-1/2)*(x^2-x+1) = 0
So let's look at (x^2-x+1) = 0
Complete the square: (x^2-x+1/4)+3/4 = 0
So we get (x-1/2)^2+3/4 = 0 or (x-1/2)^2 = -3/4
Take the square root of both sides: x-1/2 = +-sqrt(-3/4)
Leaving us with x = 1/2 +- i*sqrt(3)/2

QED

(I kind of miss algebra now)

(take a course in it you'll stop missing it p quickly)