SOLUTION: I'm having trouble with a question in my lesson. Could I have some help? (-1 + [SQRT(3)]i)^3 These are the options: A) 2 + 2[SQRT(3)]i B) -2

SOLUTION: I'm having trouble with a question in my lesson. Could I have some help? (-1 + [SQRT(3)]i)^3 These are the options: A) 2 + 2[SQRT(3)]i B) -2 - 2[SQRT(3)]i C) 0 D) -

Question 740092: I'm having trouble with a question in my lesson. Could I have some help?
(-1 + [SQRT(3)]i)^3
These are the options:
A) 2 + 2[SQRT(3)]i
B) -2 - 2[SQRT(3)]i
C) 0
D) -8
E) -8 + 8[SQRT(3)]i
F) 8
Answer by KMST(5328) (Show Source): You can put this solution on YOUR website!
You can calculate as you would calculate the cube of any binomial, using the formula

If you already learned about the polar form of complex numbers, you could transform the number into its polar form, if that made it easier for you.
has
absolute value (or modulus):
argument:
Multiplication and powers are easier in polar form.
To multiply, you just multiply the absolute values and add the arguments.
So
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