A couple of ways. This is how I would do it:

z^6 + 64 = 0

This is the sum of two perfect cubes, (z^2)^3 and 4^3 so this factors:

(z^2 + 4)(z^4 - 4z^2 + 16) = 0

So set both factors equal to 0:

z^2 + 4 = 0 gives z = (+/-) 2i

The second factor gives:

z^4 - 4z^2 + 16 = 0, which can be "solved" via the quadratic formula:

z^2 = 2 (+/-) 2i*sqrt(3)

The best way to continue from here is to convert to complex polar coordinate form:

2 + 2i*sqrt(3) = 4 exp(i*pi/3)

2 - 2i*sqrt(3) = 4 exp(-i*pi/3)

So taking the square root of both of these gives:

z = (+/-) 2 exp(i*pi/6) and z = (+/-) 2 exp(-i*pi/6)

Or if you want these in "rectangular" form:

z = (+/-) (sqrt(3) + i) and z = (+/-)(sqrt(3) - i)

(And don't forget the z = (+/-) 2i)

-Dan