Precalc, so I'll make no assumptions beyond an understanding of binomial expansion, and that i²=-1, and avoid the trigonometry entirely.

8 = 8 + 0i = (a+bi)³ = a³+3a²bi-3ab²-b³i = a³-3ab² + (3a²b-b³)i = a(a²-3b²) + b(3a²-b²)i

so a(a²-3b²) = 8 and b(3a²-b²) = 0, so b=0 so a=2, or 3a²-b² = 0, so b²=3a²,

hence after substitution a(a²-3b²) = 8 becomes a(a²-3(3a²)) = 8 so -8a³=8, so a=-1 and b = ±√3

So the solutions to (a+bi)³ = 8 are 2 + 0 i, -1 + √3 i, -1 - √3 i in the form a + b i

In the complex plane, 8 has magnitude (or length or modulus) 8 and angle 0 degrees or alternatively 360 or 720 degrees (we are considering 3 possible angles since there are 3 cube roots).

Taking a cube root results in taking the cube root of the magnitude and dividing the angle by 3.

All the cube roots have magnitude cuberoot(8) = 2, and their angles are 0 degrees, 120 degrees, and 240 degrees.

The cube roots are

2[cos (0 degrees) + i sin (0 degrees)] = 2,

2[cos (120 degrees) + i sin (120 degrees)] = 2[-1/2 + i sqrt(3)/2] = -1 + i sqrt(3).

2[cos (240 degrees) + i sin (240 degrees)] = 2[-1/2 - i sqrt(3)/2] = -1 - i sqrt(3).

The cube roots of 8 are 2, -1 + i sqrt(3), and -1 - i sqrt(3).

Have a blessed, wonderful day!

http://math.ucsd.edu/~wgarner/math4c/textbook/chap...

Scroll down to example 4. (Also read the rest, don't just copy the answer!)

what's the question?

2, 2(cos(2π/3) + i sin(2π/3)), 2(cos(4π/3) + i sin(4π/3))

2, 2(-1/2 + i √(3)/2), 2(-1/2 - i √(3)/2)

2, -1 + i √(3), -1 - i √(3)