$$\frac{1}{3c}+\frac{1}{3} = \frac{2}{c^2-4c}$$
Answer
$$ \begin{matrix}c_1 = 0 & c_2 = 4.26697 & c_3 = -0.63349+1.00242i \\[1 em] c_4 = -0.63349-1.00242i & \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} \frac{1}{3c}+\frac{1}{3} &= \frac{2}{c^2-4c}&& \text{multiply ALL terms by } \color{blue}{ 3c(c^2-4c) }. \\[1 em]3c(c^2-4c)\cdot\frac{1}{3c}+3c(c^2-4c)\cdot\frac{1}{3} &= 3c(c^2-4c)\cdot\frac{2}{c^2-4c}&& \text{cancel out the denominators} \\[1 em]c^4-4c^3+c^3-4c^2 &= 6c&& \text{simplify left side} \\[1 em]c^4-3c^3-4c^2 &= 6c&& \text{move all terms to the left hand side } \\[1 em]c^4-3c^3-4c^2-6c &= 0&& \\[1 em] \end{aligned} $$
In order to solve $ \color{blue}{ x^{4}-3x^{3}-4x^{2}-6x = 0 } $, first we need to factor our $ x $.
$$ x^{4}-3x^{3}-4x^{2}-6x = x \left( x^{3}-3x^{2}-4x-6 \right) $$$ x = 0 $ is a root of multiplicity $ 1 $.
The remaining roots can be found by solving equation $ x^{3}-3x^{2}-4x-6 = 0$.
This polynomial has no rational roots that can be found using Rational Root Test.
Roots were found using qubic formulas.
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