Need help on Pre-Calc Math?

Question

Solve the given equations in the complex number system.

x^6=-1
x^3=i
x^3+27i=0

Dont just tell me answers, tell me how to do problems like these.
tyvm

Answer 1

A key is to understand that i = √-1, so i^2 = -1, i^3 = -1 * i = -i, and i^4 = -1^2 = 1. Further exponents of i repeat the cycle.

1.) x^6=-1
√x^6 = √-1
x^3 = i (conveniently, that's #2 :)

Remember that i^3 = -i:
i^3 = -i
-i^3 = i = x^3
-i^3 = x^3
x = -i.

3.) x^3+27i=0
x^3 = -27i

Again, remember that i^3 = -i:

x^3 = 3^3 * i^3
x = 3i (solution)

Answer 2

not sure approximately all of them yet for the 1st one it appears that evidently such as you wanna do the (cube root of x-2) = 8 so which you're able to take the cube root of the two factors leaving you with: x-2 = 2 because of the fact the cube root of 8 is two. (2 * 2 * 2 = 8) upload 2 to the two factors and you have x=4