How is N^2/3 equivalent to 1/(N^1/3)? I've tried to look for similar things on StackExchange and elsewhere on the net, but can't seem to find anything, so thought I'd just ask for some help on here...
Someone has kindly helped me with a model for my research, outlined here: Probability of spherical particles touching one another in a cylindrical column
Basically, I have an equation like so:
$$\begin{aligned}
        \tau & = \frac{2f(6N-0.926N^{2/3})}{N} \\
      \tau & = 2f(6-0.926N^{-1/3})
      \end{aligned}$$
I appreciate that the 2 $N$'s cancel each other out at the beginning of the equation, but I don't understand how $N^{2/3}$ reduces to $N^{-1/3}$... is there some sort of maths rule that I'm missing or something really simple that I'm overlooking?
Any help would be greatly appreciated; thank you so much :)
 A: Yes, there's a basic rule that you're missing, and it's a pretty useful one. 
When you have $\frac{N^a}{N^b}$, you can simplify to $N^{a-b}$. In your case, $a = \frac{2}{3}$ and $b = 1$, so the final exponent is $\frac{2}{3} - 1 = -\frac{1}{3}$. 
If you're wondering why the rule is true, it's probably easiest to see when $a$ and $b$ are integers. If you have $a$ copies of $N$ on top and $b$ on the bottom, and $a \ge b$,  then you can cancel them pairwise until all the ones on the bottom are gone, and there are $a-b$ left on top. Try this with $a = 4, b = 2$ to see. 
If $a < b$, the same idea works, but you end up with $b-a$ of them on the bottom, which is $\frac{1}{N^{b-a}} = N^{-(b-a)} = N^{a-b}$ again. 
What about if they're fractions? I'll show you why it works in the case where the denominators on the fractions are the same, as in 
$$
u = \frac{N^{\frac{2}{3}}}{N^{\frac{1}{3}}}
$$
In this case, we can look at $u^3$ and see that 
\begin{align}
u^3 &= \left(\frac{N^{\frac{2}{3}}}{N^{\frac{1}{3}}}\right)^3\\
&= \frac{\left(N^{\frac{2}{3}}\right)^3}{\left(N^{\frac{1}{3}}\right)^3}\\
&= \frac{N^2}{N^1}\\
&= N
\end{align}
so that $u = N^\frac{1}{3}$ as the formula predicts. 
When the denominators are different, you can first rewrite the fractions by putting them over a common denominator, but I'm not going to work through all the details here. From the examples I've given, you're probably already starting to believe that the general rule I cited was valid. 
By the way, proving the rule when the exponents cannot be written as fractions -- when they're numbers like $\pi$ -- actually involves some relatively subtle mathematics. But then, even clearly defining $N^\pi$ requires some subtle stuff as well. 
A: Hint: 
$$
\tau=\dfrac{2f(aN-bN^{\frac{2}{3}})}{N} =\dfrac{2faN}{N}-\dfrac {2fbN^{\frac{2}{3}}}{N} = 2fa -2fbN^{(\frac{2}{3}-1)}=
$$
$$
=2fa -2fbN^{(-\frac{1}{3})}
$$
