# Simple question about nth-roots of negative numbers

1. Feb 9, 2012

### GCH

I've been doing this online math bridging course from a German uni to which I am planning to go soon, it is going through all high school maths from the beginning(it has been a while since I did it).
The chapter on roots so far ignores complex numbers, which will be introduced in a later chapter... but they claim that $\sqrt[3]{-125}$ = -5 ! Can this be correct? Surely not?
I always thought that any root of a negative number has only complex solutions (I know that positive ones have complex and real solutions)
What is $\sqrt[3]{-1}$ anyway? (I am having a bit of a brain-freeze at the moment.)

Thank you!

2. Feb 9, 2012

### Char. Limit

Every number has THREE cube roots. Of these three, one will be real, and the other two will be complex conjugates of each other. And yes, every real number has one real cube root. $\sqrt[3]{-1} = -1$, but that's only one of the three solutions to x^3 = -1.

3. Feb 9, 2012

### SteveL27

What is -1 * -1 * -1?

4. Feb 9, 2012

### HallsofIvy

Surely yes. $(-5)^3= (-5)(-5)(-5)= (25)(-5)= -125$

Any even root of a negative number has only complex solutions. (Strictly speaking, you mean "non-real", not complex. The real numbers are a subset of the complex numbers.)

As SteveL27 said, $(-1)^3= (-1)(-1)(-1)$ so one third root of -1 is -1. The other two are $1/2+ i\sqrt{3}/2$ and $1/2- i\sqrt{3}/2$.

Last edited by a moderator: Feb 10, 2012
5. Feb 10, 2012

### GCH

Thank you for clearing this up for me. Somehow I did not think last night of asking myself the question: what number multiplied an uneven number of times yields -1 as a result.