Given semiperimeter and radii of inscribed and circumscribed circles, find the side lengths of triangle Consider $\triangle ABC$ with side lengths $a,b,c$,
semiperimeter $\rho=\tfrac12\,(a+b+c)$,
inradius $r$ and circumradius $R$.
Let $u=\rho/R,\ v=r/R$,
$a'=a/R,\ b'=b/R,\ c'=c/R$.
Given $u,\,v$,
we can find 
$a',\,b',\,c'$ as roots of cubic equation
\begin{align}
x^3-2\,u\,x^2+(u^2+v^2+4\,v)\,x-4\,u\,v&=0
\tag{1}\label{1}
,
\end{align}
and the side lengths in increasing order $a\le b\le c$
can be expressed as
\begin{align}
a&=
\tfrac23\,R\,\left(u+\sqrt{u^2-3\,v\,(v+4)}\,
    \cos(\phi+\tfrac{2\pi}3)
  \right)
\tag{2a}\label{2a}
,\\
b&=
\tfrac23\,R\,\left(u+\sqrt{u^2-3\,v\,(v+4)}\,
    \cos(\phi-\tfrac{2\pi}3)
  \right)
\tag{2b}\label{2b}
,\\
c&=
\tfrac23\,R\,\left(u+\sqrt{u^2-3\,v\,(v+4)}\,
    \cos(\phi)
  \right)
\tag{2c}\label{2c}
,\\
\phi&=
      \tfrac13\,\arctan\left(
        {3\,\sqrt3\,v\,\sqrt{u^2\,(2\,(27-(5-v)^2)-u^2)-v\,(v+4)^3}}
        ,
        {u\,(9\,v\,(2-v)-u^2)}
      \right)
\tag{2phi}\label{2phi}
.
\end{align}
Numerical tests demonstrate that in general 
expressions \eqref{2a}-\eqref{2c}
work fine for the valid range of $u,\,v$, 
except for
an annoying special case, 
which corresponds to equilateral triangle,
when $v=\tfrac12$, $u=\tfrac{3\sqrt3}2$,
$a=b=c=\tfrac23\,u\,R$.
In this case \eqref{1} collapses to 
\begin{align}
(x-\sqrt3)^3&=0
\tag{3}\label{3}
\end{align}
and direct numeric calculation of $\phi$ fails.
Of course, it's not a big deal and this case can be algorithmically isolated and handled differently,
but,
the question is: Is it possible to come up with some modified version
of expression for $\phi$, which would work
gracefully for all valid pairs of $u,\,v$?
This probably could be useful:
the valid range is found as follows: for $v\in(0,\tfrac12]$,
\begin{align}
u&\in[u_{\min},u_{\max}]
,\\
u_{\min}(v)&=
\sqrt{27-(5-v)^2-2\,\sqrt{(1-2\,v)^3}}
,\\
u_{\max}(v)&=
\sqrt{27-(5-v)^2+2\,\sqrt{(1-2\,v)^3}}
.
\end{align} 
Both boundary curves $u_{\max}(v)$ and $u_{\min}(v)$, 
correspond to isosceles triangles,
$u_{\max}(v)$ represents all isosceles triangles 
with the base as the smallest side,
while 
$u_{\min}(v)$ represents all isosceles triangles 
with the base as the largest side.
The area, bounded by $u_{\min}(v)$, $u_{\max}(v)$
and the $u$-axis covers all possible types of valid triangles
with $R=1$ and can be used as a "map of the Trianglia"
to make a quick crude tests of some general properties of triangles,
for example, to estimate $\max(\sin\alpha+\sin\beta+\sin\gamma)$, etc.
 A: (Converting comment to answer, as requested.)
Triangles aside, you are effectively asking for a universal solution to the (depressed) cubic $t^3+pt+q=0$ that works even when $p=q=0$. I'm personally not aware of one. Solutions such as those in Wikipedia separate-out such special cases. (Of course, when $p=0$, the solution is clear enough not to need a "formula".) 
For completeness: Your cubic's corresponding "depressed" coefficients are
$$\begin{align}
p &=\frac13 (3v^2+12 v-u^2) = \frac1{3R^2}\left(3r^2+12\rho R−\rho^2\right) \\[6pt]
&=−\frac1{3R^2}\left(a^2+b^2+c^2−ab−bc−ca\right) \\[10pt]
q 
&=\frac2{27} u \left(u^2 - 18 v + 9 v^2\right) = \frac2{27R^3}\rho\left(9r^2−18rR+ρ^2\right) \\[6pt]
&=−\frac1{27R^3}(2a−b−c)(−a+2b−c)(−a−b+2c)
\end{align}$$
which (obviously) vanish when $a=b=c$.
Note that (a multiple of) $\sqrt{-p}$ is the coefficient of the $\cos(\phi+\cdots)$ terms in OP's expressions for the roots. So, maybe the fact that the $\phi$ formula fails when $p=0$ doesn't really matter, since those terms simply vanish.
A: EDIT1:
Not an answer really. As an aside like to how geometrically how $ (R,r,\rho?,a,b,c) $ can be represented by a  polar diagram from a cubic.
Mathematica outputs a big analytical solution.. may be relatable to irreducible case of Cardano's solution of the cubic equation.. that yields three real roots.
$ (R,r,s) = (4.04145, 0.866025, 7.5)$ numerical values belong to an obtuse angled triangle of sides and semi-perimeter $(a= 7, b= 5,c= 3, \rho=7.5)$
