# ((-8)^2)^(1/3) <> (-8)^(2/3) <> (-8)^(1/3))^2

1. Feb 10, 2016

((-8)^2)^(1/3)=64^(1/3)=4
(-8)^(1/3))^2 = (-2)^2=4
but
(-8)^(2/3) = 4(-1)^(2/3) = 4((2/3)(cos (π) + i*sin(π)) and so forth, giving a complex number.
I understand why the latter one works. Why don't the first two?
Thanks.

2. Feb 10, 2016

### maNoFchangE

Why does 2/3 act as a multiplicative constant instead of an exponential? Moreover $\sin \pi =0$.

3. Feb 10, 2016

### Staff: Mentor

4. Feb 10, 2016

Oops, too quickly typed the last equation. I meant to multiply the two-thirds times the angles, not times the whole expression (or equivalently use it as an exponent), using Euler's formula. Anyway, my question is not about Euler's formula, which as I say I do get when I am not typing while my bus is about to leave, but rather why the first two formulas are invalid.

5. Feb 10, 2016

### HallsofIvy

No, it doesn't. sin(π)= 0. The other two cube roots are non-real, of course.
The first two do also, you just have restricted yourself to real numbers. 64 has three complex cube roots, 4 is one, the other two are non-real. 8 has 3 cube roots, one is 2 and the other two are non-real.

6. Feb 10, 2016

### Staff: Mentor

They are valid. They give you one of possible three answers, the real one.

7. Feb 11, 2016