How to show that the roots of $-x^3+3x+\left(2-\frac{4}{n}\right)=0$ are real (and how to find them) I'm trying to find the three distinct and real roots of 
$$-x^3+3x+\left(2-\frac{4}{n}\right)=0,$$
where $n>0$ (we could say $n\geq 2$ if that helps), but I'm not able to get very far:
Using the notation of the Wikipedia-page, I find that the discriminant is 
$$\Delta=27\cdot 4\left(\frac{1}{n}-\frac{1}{n^2}\right),$$
which gives 
$$C=3\left( 1-\frac{2}{n} \pm 2 i \sqrt{\frac{1}{n}-\frac{1}{n^2}} \;\right)^{1/3},$$
which is then used to find $x$, as 
$$x=\frac{C}{3}+\frac{3}{C}.$$
Two things confuse me: 


*

*$C$ looks like a non-real number, but $x$ should be real (since $\Delta >0$). How can one further reduce the expression for $x(C)$ to show that the imaginary part is zero? I'm having trouble evaluating that cube root.

*When I use my expression for $x(C)$ in Mathematica and evaluate numerically for some $n$, I only find one of the three solutions that Mathematica finds if I just ask it to give the roots of the original equation (the change of sign in $C$, i.e. the $\pm$, doesn't even give two different solutions). What have I done to exclude the two other solutions (or is it just Mathematica excluding them somehow)?


Context:
Actually, I'm only trying to find the root where $-1\leq x\leq 1$. I'm trying solve (the first) part of system of equations that I solved numerically in order to make this answer analytically, so that I can play with the limit of the expression (as $n\rightarrow \infty$). 
Thank you. 

For the calculation of the roots of the depressed cubic
$$
y^{\,3}  + p\,y + q = 0
$$
where $p$ and $q$ are real or complex,
I personally adopt a method indicated in this work by A. Cauli, by which putting
$$
u = \sqrt[{3\,}]{{ - \frac{q}
{2} + \sqrt {\frac{{q^{\,2} }}
{4} + \frac{{p^{\,3} }}
{{27}}} }}\quad v =  - \frac{p}
{{3\,u}}\quad \omega  = e^{\,i\,\frac{{2\pi }}
{3}} 
$$
where for the radicals you take one value, the real or 
the first complex one (but does not matter which)
then you compute the three solutions as:
$$
y_{\,1}  = u + v\quad y_{\,2}  = \omega \,u + \frac{1}
{\omega }\,v\quad y_{\,3}  = \frac{1}
{\omega }\,u + \omega \,v
$$
In your case:
$$
y^{\,3}  - 3\,y - 2\left( {\frac{{n - 2}}
{n}} \right) = 0
$$
we obtain
$$
\frac{{q^{\,2} }}
{4} + \frac{{p^{\,3} }}
{{27}} = \left( {\frac{{n - 2}}
{n}} \right)^{\,2}  - 1 =  - 4\frac{{\left( {n - 1} \right)}}
{{n^{\,2} }} < 0
$$
which confirms that there are three real solutions, and
$$
\begin{gathered}
  u = \sqrt[{3\,}]{{\frac{{n - 2}}
{n} + i\,\frac{2}
{n}\sqrt {\left( {n - 1} \right)} }} = \frac{1}
{{\sqrt[{3\,}]{n}}}\;\sqrt[{3\,}]{{n - 2 + i\,2\sqrt {\left( {n - 1} \right)} }} =  \hfill \\
   = \frac{1}
{{\sqrt[{3\,}]{n}}}\;\sqrt[{3\,}]{{n\,e^{\,i\,\alpha } }} = e^{\,i\,\alpha /3} \quad \left| {\,\alpha  = \arctan \left( {\frac{{2\sqrt {\left( {n - 1} \right)} }}
{{n - 2}}} \right)} \right. \hfill \\
  v =  - \frac{p}
{{3\,u}} = \frac{1}
{u} = e^{\, - \,i\,\alpha /3}  \hfill \\ 
\end{gathered} 
$$
with the understanding that for $n=1,\; 2$,  $\alpha= \pi , \; \pi /2$, i.e. that we use the 4-quadrant $arctan$.
So that in conclusion, for $0<n$, we have:
$$
\left\{ \begin{gathered}
  y_{\,1}  = e^{\,i\,\alpha /3}  + e^{\, - \,i\,\alpha /3}  = 2\cos \left( {\frac{\alpha }
{3}} \right) \hfill \\
  y_{\,2}  = e^{\,i\,\alpha /3 + 2\pi /3}  + e^{\, - \,i\,\alpha /3 - 2\pi /3}  = 2\cos \left( {\frac{{\alpha  + 2\pi }}
{3}} \right) \hfill \\
  y_{\,3}  = e^{\,i\,\alpha /3 - 2\pi /3}  + e^{\, - \,i\,\alpha /3 + 2\pi /3}  = 2\cos \left( {\frac{{\alpha  - 2\pi }}
{3}} \right) \hfill \\ 
\end{gathered}  \right.
$$
Concerning the range spanned by the solutions, apart for $n=1$ where we get the solutions (1,-2,1), then
for $2 \le\; n$ we have
$$
\frac{{\alpha (n)}}
{3}\quad \left| {\;2 \leqslant n} \right.\quad  = \frac{1}
{3}\arctan _{\,4\,Q} \left( {n - 2,\;2\sqrt {\left( {n - 1} \right)} } \right) = \left\{ {\frac{\pi }
{6},\frac{\pi }
{{7.66}},\; \cdots } \right\}
$$
which means:
$$
\left\{ \begin{gathered}
  \quad \quad 2 \leqslant n \hfill \\
  0 < \frac{{\alpha (n)}}
{3} \leqslant \frac{\pi }
{6}\quad  \Rightarrow \quad \sqrt 3  \leqslant y_{\,1}  < 2 \hfill \\
  2\frac{\pi }
{3} < \frac{{\alpha (n)}}
{3} + 2\frac{\pi }
{3} \leqslant \frac{5}
{6}\pi \quad \quad  \Rightarrow \quad  - 2 < y_{\,2}  \leqslant  - \sqrt 3  \hfill \\
   - 2\frac{\pi }
{3} < \frac{{\alpha (n)}}
{3} - 2\frac{\pi }
{3} \leqslant  - \frac{\pi }
{2}\quad \quad  \Rightarrow \quad  - 1 < y_{\,3}  \leqslant 0 \hfill \\ 
\end{gathered}  \right.
$$
