# 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.

• Is $n$ a natural number ? – Debashish Jun 27 '14 at 8:45
• The cubing on the LHS is redundant and serves no purpose, so you can simplify life by ignoring it. For $x^{3}=0$ if and only if $x = 0$ when $x$ is a real or complex number. – Geoff Robinson Jun 27 '14 at 8:52
• @GeoffRobinson: Or simply in any field, since no non-zero elements have product being zero. – user21820 Jun 27 '14 at 8:54
• Yes, indeed, in any field, but I was not sure how many fields the questioner might have seen. I just wanted to be careful, because the statement is not true in some rings, eg, rings of matrices. – Geoff Robinson Jun 27 '14 at 8:56
• Can you enplane why the cubing is redundant? – Ghost waffles Jun 27 '14 at 9:02

## 2 Answers

$(\frac{2}{(n^{-2}-n^{-3})}+1)^3=0$

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

• I mean multiply both sides of the fraction, i.e. the numerator and denominator – Wonder Jun 27 '14 at 10:54
• Oh, I see :) linguistic misunderstanding. – rschwieb Jun 27 '14 at 11:27

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.

• What if n is not a real number? @Thoth19 – Ghost waffles Jun 27 '14 at 22:54
• Well then that complicates things. What is n? Maybe it is in $\mathbb Z /5$ – Thoth19 Jun 28 '14 at 0:25