Express a formula in terms of trigonometric expressions I am studying Kerr Black holes using Hobson's General relativity an introduction for physicists book. 
In order to find circular radius for photons, two conditions need to be satisfied:
$$r_c=3\mu\frac{b-a}{b+a}$$ and
$$(b+a)^3=27\mu^2(b-a)$$
According to the book: The equations may be solved by setting y=a+b in the second condition and substituting the resulting value of b into the first. This is what I did and I obtained:
$$r_c=\frac{\mu\alpha^{1/3}(3\mu^2+3\alpha^{1/3}-2a)}{\mu^2+\alpha^{2/3}}$$
where for simplification I defined $\alpha=\sqrt{a^2\mu^4-\mu^6}-a\mu^2$.
However, the book further says that one can easily simplify $r_c$ as:
$$r_c=2\mu\Big(1+\cos\Big[\frac{2}{3}\cos^{-1}\Big(\pm\frac{a}{\mu}\Big)\Big]\Big)$$
and $$b=3\sqrt{\mu{r_c}}-a$$
I am stuck and don't know how is it possible to simplify my expression for $r_c$ into the elegant expression as given by the book. This seems to be a physics question but I am stuck with the algebraic manipulation.

\begin{align}
r_c&=3\mu\frac{b-a}{b+a} \tag{1}\label{1}
\\
(b+a)^3&=27\mu^2(b-a) \tag{2}\label{2}
\end{align}
To get the expression for b from \eqref{1}, 
\begin{align}
b-a&=\frac{r_c}{3\mu}(b+a)
,
\end{align}
combined with \eqref{2},
\begin{align}
 (b+a)^3&=27\mu^2\frac{r_c}{3\mu}(b+a)
 ,\\
 (b+a)^2&=9\mu{r_c}
 ,\\
 b&=3\sqrt{\mu{r_c}}-a
 .
\end{align}
From \eqref{2}
\begin{align}
3\,\mu\,\frac{b-a}{b+a}
&=\frac{(b+a)^2}{9\,\mu}=r_c
\tag{3}\label{3}
.
\end{align}
Now we have two expressions for $r_c$.
One has a factor $\frac{\mu}{(b+a)}$,
the other has its reciprocal $\frac{(b+a)}{\mu}$,
and they both are begging to be canceled.
When we multiply them, we'll get a nice simplified 
expression for $r_c^2$:
\begin{align}
r_c^2&=
\tfrac13\,(b+a)(b-a)
,\\
r_c^2&=\sqrt{\mu\,r_c}(3\,\sqrt{\mu\,r_c}-2\,a)
,\\
\left(\frac{r_c}{\mu}\right)^2
&=
\sqrt{\frac{r_c}{\mu}}
\left(
3\,\sqrt{\frac{r_c}{\mu}}-\frac{2\,a}{\mu}
\right)
,\\
\left(\sqrt{\frac{r_c}{\mu}}\right)^3
-
3\,\sqrt{\frac{r_c}{\mu}}
&=
-\frac{2\,a}{\mu}
,\\
4\,\left(\tfrac12\sqrt{\frac{r_c}{\mu}}\right)^3
-
3\,\left(\tfrac12\sqrt{\frac{r_c}{\mu}}\right)
&=
-\frac{a}{\mu}
\end{align}
Recall that
\begin{align}
4\,\cos^3 x-3\,\cos x=\cos3x.
\end{align}
So, we have 
\begin{align}
\cos3x&=-\frac{a}\mu
;\\
3x&=\arccos\left(-\frac{a}\mu\right)+2\,\pi k,\quad k=0,1,2
;\\
x&=\tfrac13\arccos\left(-\frac{a}\mu\right)+\tfrac23\,\pi k,\quad k=0,1,2
;\\
\end{align}
Hence
\begin{align}
\cos x=
\tfrac12\sqrt{\frac{r_c}{\mu}}
&=
\cos\left(
\tfrac13\,\arccos\left( -\frac{a}{\mu} \right)
+\tfrac23\,\pi\,k
\right)
,\quad k=0,1,2
;\\
\tfrac12\frac{r_c}{\mu}
&=
2\,
\cos^2\left(
\tfrac13\,\arccos\left( -\frac{a}{\mu} \right)
+\tfrac23\,\pi\,k
\right)
,\quad k=0,1,2
;\\
r_c&=2\,\mu
\left(
1+\cos\left(
\tfrac23\,\arccos\left( -\frac{a}{\mu} \right)
+\tfrac43\,\pi\,k
\right)
\right)
,\quad k=0,1,2
.
\end{align}
