Need help with algebra $$\left(\frac{2}{n^{-2}-n^{-3}}+1\right)^3=0$$
How do I do this? I have tried for a week to solve it. 
I tried just doing simple algebra and each time I get something different or get stuck
Then some guy said i can solve it using functions so i tried that and just got lost.
 A: $(\frac{2}{(n^{-2}-n^{-3})}+1)^3=0$
Means $(\frac{2}{(n^{-2}-n^{-3})}+1)=0$
Simplify the fraction by multiplying both sides by $n^3$.
$\frac{2n^3}{n-1} = -1$.
This gives us the cubic equation $2n^3 + n - 1 = 0$.
The only real root is about   0.5897545123014584
A: First, I'm going to go out on a limb and assume that $n \in \mathbb R$. So that means, that $\frac{2}{n^{-2}-n^{-3}}+1 = 0$ because nothing to the third power is 0 apart from 0. 
Now we have to move our messy bits around $2 = -1 * (n^{-2}-n^{-3})$ which implies $-2 = (\frac{1}{n^{2}}-\frac{1}{n^{3}})=(\frac{n}{n^{3}}-\frac{1}{n^{3}})=(\frac{n-1}{n^{3}})$.
Now we take $-2n^3 = n - 1$ which implies $0=2n^3+n-1$. Now we need to take the roots. For this we will use the cubic formula (https://en.wikipedia.org/wiki/Cubic_function#General_formula_for_roots) (or just a calculator) to find that n = .5897...
We could also use some calculus, a la Newton's Method to solve for this value, but writing out that table is something I would prefer to avoid doing unless you want to see it done out. Hope this helps.
