I have the cubic equation: \begin{equation} \alpha^3 - 3\alpha + \left(2 - \frac{3}{2\eta}\right)=0 \end{equation}
This is formed such that $\alpha$ is a fraction such that $0<\alpha<1$. I am interested in solutions to this cubic equation when the parameter $\eta$ has two different ranges of values: $\eta \geq 2$, and $1<\eta<2$. The former is of primary interest.
I have solved the equation using numerical methods, and always find three real roots, although only one which satisfied the condition $0<\alpha<1$. However, it would be most useful to have an analytical expression for these roots.
Therefore, I have used Mathematica (using $a$ and $n$ for $\alpha$ and $n$) to obtain the following:
\begin{equation} \left\{\left\{a\to \frac{\sqrt[3]{\frac{\sqrt{3} \sqrt{3-8 n}}{n}+\frac{3}{n}-4}}{2^{2/3}}+\frac{2^{2/3}}{\sqrt[3]{\frac{\sqrt{3} \sqrt{3-8 n}}{n}+\frac{3}{n}-4}}\right\},\left\{a\to -\frac{\left(1-i \sqrt{3}\right) \sqrt[3]{\frac{\sqrt{3} \sqrt{3-8 n}}{n}+\frac{3}{n}-4}}{2\ 2^{2/3}}-\frac{1+i \sqrt{3}}{\sqrt[3]{2} \sqrt[3]{\frac{\sqrt{3} \sqrt{3-8 n}}{n}+\frac{3}{n}-4}}\right\},\left\{a\to -\frac{\left(1+i \sqrt{3}\right) \sqrt[3]{\frac{\sqrt{3} \sqrt{3-8 n}}{n}+\frac{3}{n}-4}}{2\ 2^{2/3}}-\frac{1-i \sqrt{3}}{\sqrt[3]{2} \sqrt[3]{\frac{\sqrt{3} \sqrt{3-8 n}}{n}+\frac{3}{n}-4}}\right\}\right\} \end{equation} Hence I have expressions for the three roots. However the quantity $3-8n$ will be inherently negative, due to the constraint that $n>1$, and so all these roots have complex components. I can see there is some form of structure in the roots but am struggling to see how to simplify this to obtain simpler expressions for the roots that do not involve $i$.
The reason I am looking to do this is more about the method for simplification. This equation results from a simple scenario and I am looking to consider more complex scenarios and to be able to compare their roots to this analytical solution (if one can be found). It looks to me like a rearrangement of these equations, but I can;t see my way through that.
Any help or comments most welcome.
I shall just follow the steps given here.
$$\alpha^3 - 3\alpha + \left(2 - \frac{3}{2\eta}\right)=0$$
You have $$\Delta=\frac{81 (8 \eta -3)}{4 \eta ^2}$$ So, three real roots if $\eta >\frac 38$.
For this case, use the trigonometric method which will give $$\alpha_k=2 \cos \left(\frac{2 \pi k}{3}-\frac{1}{3} \cos ^{-1}\left(\frac{1}{2} \left(\frac{3}{2 \eta }-2\right)\right)\right)\qquad \text{for}\qquad k=0,1,2$$
The only one that shows an $0\leq \alpha \leq 1$ is $\alpha_1$, and this is only for $\eta\ge\frac34$. Putting $\frac38<\eta<\frac34$ results in $\alpha_1$ being negative instead, and the other two roots always have absolute values greater than $1$.
For the case where $\eta<\frac 38$, one real root. Using the hyperbolic method, it would be $$\alpha=2 \text{sgn}(\eta ) \cosh \left(\frac{1}{3} \cosh ^{-1}\left(\frac{3-4 \eta }{4 |\eta |}\right)\right)$$