I am sorry for wasting your time with elementary mathematics but I can not complete my homework.

(excuse the clutter, for I do not know how to use Latex)

cube root ( -125 m^-4 n^5 )
I fully simplify this to:
-5n/m * cube root (n^2 / m )

The answer is incorrect, wtf am I missing?

Do not feel that you are wasting our time: all of us choose to visit this forum and answer questions.

I'm not sure how they want you to simplify this, but remember the following:
$$\sqrt[x]{y^{z}}=y^{\frac{z}{x}}$$

Yes, I know that, but we are to simplify them while still having the radical, but my answer is somehow wrong.
-5n/m^2 * cube root(m^2 n^2)
I can not figure out how thay came to that answer.

There are multiple correct answers here. In my opinion, if you simplify by giving m and n exponents in the form of fractions, you can't simplify much more. Both this answer and the one you gave above are correct; just different.

Thank you. I was a nervous wreck thinking I had missed something in class, or could not understand the question.

If you are ever unsure, just take two numbers and plug them in. For example, n = 1, m = 2 and then make sure you get the same answer before and after you simplify.

The problem you're having stems from the fact that you're not supposed to have radicals in the denominator. What you have to do is this to get an acceptible answer:

$$\frac{-5n}{m}\sqrt[3]{\frac{n^2} {m}}$$

Now, multiply the whole problem by 1--in disguise though:

$$(\frac{m}{m})\frac{-5n}{m}\sqrt[3]{\frac{n^2} {m}}$$

remember, m divided by m is 1.

Now, move the m in the numerator into the radical by saying: $m=\sqrt[3]{m^3}$

Thus you have:
$$\frac{-5n}{m^2}\sqrt[3]{m^3\frac{n^2} {m}}$$

and of course the m in the denominator of the radical cancels oy leaving you with the correct answer.

$$\frac{-5n}{m^2}\sqrt[3]{m^2n^2}$$

This is what you're "supposed" to do when you have a radical in the denominator. Will you do this all the time? No. One example of not bothering to do this is $\cos 45^\circ=\frac{1}{\sqrt{2}}$

You'll see these trig values written with the radical in the denominator quite often because the radical can be cancel later on or the required answer will be a decimal approximation.

Hope this helped.

Requirements for radicals in the denominator vary from program to program. In Math 31 IB, we were told not to bother. In other courses, we were not. Ultimately, the answer is correct just in a different form. Nevertheless, faust has answered uranium's question as to how they got to that answer.

uranium_235 said:
(excuse the clutter, for I do not know how to use Latex)
It is good to practise LaTeX. Many maths symbols and Greek characters can be written :
$$\tau\upsilon\rho\nu\beta\alpha\chi\kappa\sigma\iota\gamma\alpha\nu\delta\epsilon\lambda\iota\mu\iota\nu\alpha\tau\epsilon\tau\eta\epsilon\rho\epsilon\pi\epsilon\alpha\tau\iota\nu\gamma\kappa\epsilon\psi$$