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. 
 A: 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:
$$
\left\{ \begin{gathered}
  \quad \quad 2 \leqslant n \hfill \\
  0 < \frac{{\alpha (n)}}
{3} \leqslant \frac{\pi }
{6}\quad  \Rightarrow \quad \sqrt 3  \leqslant y_{\,1}  < 2 \hfill \\
  2\frac{\pi }
{3} < \frac{{\alpha (n)}}
{3} + 2\frac{\pi }
{3} \leqslant \frac{5}
{6}\pi \quad \quad  \Rightarrow \quad  - 2 < y_{\,2}  \leqslant  - \sqrt 3  \hfill \\
   - 2\frac{\pi }
{3} < \frac{{\alpha (n)}}
{3} - 2\frac{\pi }
{3} \leqslant  - \frac{\pi }
{2}\quad \quad  \Rightarrow \quad  - 1 < y_{\,3}  \leqslant 0 \hfill \\ 
\end{gathered}  \right.
$$
A: 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.
A: $C/3$ has absolute value $1$, so $3/C=(C/3)^{-1}=\overline{C/3}$, therefore $x=(C/3)+\overline{C/3}=2\operatorname{Re}(C/3)$ is real.

Since $(C/3)^3=(1-\frac{2}{n})\pm i\cdot(2\sqrt{\frac{1}{n}-\frac{1}{n^2}})$, we use the definition of absolute values to obtain
$$|C/3|^6=|(C/3)^3|^2=(1-\frac{2}{n})^2+4\left(\frac{1}{n}-\frac{1}{n^2}\right)=1$$
and therefore $|C/3|=1$.
As to how to calculate $C$ and $x$ more directly: In the definition of square root we need there is a cubic root, so we need to write $C$ in polar form: Say $C/3=e^i\theta$. The sign of $\pm$ in the definition of $C$ will be the same as the sign of $\theta$.
Then $x=C/3+3/C=2\operatorname{C/3}=2\cos(\theta)$. Since $\cos$ is even, the sign of $\theta$ (which is the sign we choose when calculating $C$) doesn't matter.
Now to calculate $\theta$: Looking at the real and imaginary parts of $1-\frac{2}{n}+2i\sqrt{\frac{1}{n}-\frac{1}{n^2}}$, we obtain
$$C/3=\left(e^{i\arctan\left(\frac{2\sqrt{\frac{1}{n}-\frac{1}{n^2}}}{1-\frac{2}{n}}\right)}\right)^{1/3}=\left(e^{i\arctan\left(\frac{2\sqrt{n-1}}{n-2}\right)}\right)^{1/3}=e^{i\frac{\arctan\left(\frac{2\sqrt{n-1}}{n-2}\right)}{3}}$$
so $\theta=\frac{\arctan\left(\frac{2\sqrt{n-1}}{n-2}\right)}{3}$ works, in which case $x=2\cos\frac{\arctan\left(\frac{2\sqrt{n-1}}{n-2}\right)}{3}$.

As for the other roots. Denote by $p(x)$ the initial polynomials. The derivative of the polynomials is $-3x^2+3$, so maximums and minimums are attained at $\pm 1$.
So $p(x)$ starts (for "very negative" numbers) large, and decreases until it gets to the first local minimum, on which
$$p(-1)=-(-1)^3+3(-1)+\left(2-\frac{4}{n}\right)=-\frac{4}{n}$$
so it must have passed through a root at some point.
After $-1$, $p(x)$ starts increasing until it gets to the local maximum at $x=1$, on which
$$p(1)=-1^3+3(1)+\left(2-\frac{4}{n}\right)=4-\frac{4}{n}\geq 0$$
so there will be another root in $(-1,1]$ (more precisely, $1$ will be a double root iff $n=1$, and if $n\neq 1$ there will be another root strictly in $(-1,1)$).
After that, $p(x)$ decreases all the way to $-\infty$, in which case it will necessarily pass through another root (if $n\neq 1$).
In the end, this shows that $p(x)$ has three real roots. One of them is found by the procedure above, and after we calculate it we can simply decompose $p(x)$ as a product of a linear and a quadratic, and solve the quadratic.
