Question

Solve the equation in the complex number system. x^3-1=0

Answer 1

I got it down to (x-1)(x^2+1x+1), but I don't know how to finish it.

Answer 2

ok... so thats great you recongnised the difference of 2 cubes

now use the general quadratic formula for the quadratic factor

\[x^2 + x + 1\]

means a = 1, b = 1 and c=1

substitute them into

\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]

you'll have a negative value in the square root...

and use

\[i^2 = -1 \]

to manage it and then you'll be able to find the complex solutions.

hope it helps

Answer 3

Oh, duh! That makes perfect sense. Thank you. :)