Find the cube roots of -64 of set of complex numbers .write the result in rectangular form a+bi

Answer 1

one cube root = -4

(-4)^3 = -64

x^3 =-64

x^3 -+64 = 0 factor to get

= (x+4)(x^2 - 4x+-16) = 0 (if you can't see the factors or know the formula, divide x+4 into x^3+64)

x^2 -4x +16 = 0 complete the square

x^ -4x + 4 = -16 +4 =- 12

(x-2)^2 = -12 take square roots of both sides

x-2 = sqr(-12)

x = 2 + or - sqr(-12)

x = 2 + 2isqr3 are two imaginary cube roots of -64

in a+bi form the non-real complex roots of -64

are 2 + 2sqr(3)i and 2- 2sqr(3)i

the 3rd cube root in a+bi form

is -4 + (0)i