Question 740092
You can calculate {{{(-1+sqrt(3)i)^3}}} as you would calculate the cube of any binomial, using the formula
{{{(a+b)^3=a^3+3a^2b+3ab^2+b^3}}}
{{{(-1+sqrt(3)i)^3=(-1)^3+3(-1)^2sqrt(3)i+3(-1)(sqrt(3)i)^2+(sqrt(3)i)^3= -1+3sqrt(3)i+3(-1)(sqrt(3))^2i^2+(sqrt(3))^3i^3= -1+sqrt(3)i+3(-1)(3)(-1)+(3sqrt(3))(-i)= -1+3sqrt(3)i+9-3sqrt(3)i=8}}}
 
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.
{{{-1+sqrt(3)i=2(cos(120^o)+i*sin(120^o))}}} has
absolute value (or modulus): {{{r=2 =sqrt((-1)^2+(sqrt(3))^2)}}}
argument: {{{theta=120^o}}}
Multiplication and powers are easier in polar form.
To multiply, you just multiply the absolute values and add the arguments.
So {{{(2(cos(120^o)+i*sin(120^o)))^3=2^3(cos(3*120^o)+i*sin(3*120^o))=8(cos(360^o)+i*sin(360^o))=8(1+i*0)8*1=8}}}