How to show that the roots of $-x^3+3x+\left(2-\frac{4}{n}\right)=0$ are real (and how to find them)
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:
$$
\left\{ \begin{gathered}
\quad \quad 2 \leqslant n \hfill \\
0 < \frac{{\alpha (n)}}
{3} \leqslant \frac{\pi }
{6}\quad \Rightarrow \quad \sqrt 3 \leqslant y_{\,1} < 2 \hfill \\
2\frac{\pi }
{3} < \frac{{\alpha (n)}}
{3} + 2\frac{\pi }
{3} \leqslant \frac{5}
{6}\pi \quad \quad \Rightarrow \quad - 2 < y_{\,2} \leqslant - \sqrt 3 \hfill \\
- 2\frac{\pi }
{3} < \frac{{\alpha (n)}}
{3} - 2\frac{\pi }
{3} \leqslant - \frac{\pi }
{2}\quad \quad \Rightarrow \quad - 1 < y_{\,3} \leqslant 0 \hfill \\
\end{gathered} \right.
$$
This is an approach on how to solve the DEPRESSED cubic equation. To show the roots are real just find the discriminant of the cubic, (it can be shown easily)
To find the roots of your query follow these steps:
Multiply orginal equation by -1.
To get $$x^3 -3x+(\frac{4}{n}-2) =0$$
Subsitute $$x= ucosa$$
Substite in the original equation to get
$$(ucosa)^3 -3ucosa+(\frac{4}{n}-2) =0$$
Recall the identity $$cos(3a)=4cos^3a-3cosa$$
Manipulate the equation to get
$$u^3(\frac{cos3a +3cosa}{4}) -3ucosa+(\frac{4}{n}-2) =0$$
Distribute to obtain: $$u^3(\frac{cos3a}{4})+ u^3(\frac{3cosa}{4}) -3ucosa+(\frac{4}{n}-2) =0$$
Take the common factor out of the middle two terms to obtain:
$$u^3(\frac{cos3a}{4})+ 3ucosa(\frac{u^2}{4}-1)+(\frac{4}{n}-2) =0$$
"u" is a parameter chosen by us. We are in search of such a parameter "$u$"; so that the middle term cancels out so that we can solve the trigonometric equation. In order for this to occur: $$u=±2$$
Substituting $u=2$ we get
$$2cos3a+(\frac{4}{n}-2) =0$$
Solving for $cos3a$ we get
$$cos3a= 1- \frac{2}{n}$$
Therefore
$$3a= arccos(1- \frac{2}{n})$$
$$a = \frac{arccos(1- \frac{2}{n})}{3}$$
$$x= 2cos(\frac{arccos(1- \frac{2}{n})}{3})$$
Similarly we can subsitute $u=-2$ and find another root.
You will get $$x=-2cos(\frac{arccos(\frac{2}{n}-1)}{3})$$
Here is a link Wiki
In fact an general solution is for the cubic equation you have demanded is:
$$x=2cos(\frac{arccos(1- \frac{2}{n})-2πk}{3})$$
Where k = 0,1,2.
Hope this helped.