Solve two equations for $a$ and $b$ \begin{cases}
c_2=\dfrac{c_1}{a} \left( \left(\dfrac{c_3}{b}\right)^3 - 1 \right) \\[2ex]
b^2 = a^2 + c_3^2 + 2(a)\, (c_3)\, (c_4)  \\
\end{cases}
I am stuck at this point. Not sure on how to move forward.  ( A small change made)

The system \begin{cases}
c_2=\dfrac{c_1}{a} \left( \left(\dfrac{c_3}{b}\right)^3 - 1 \right) \\[2ex]
b^2 = a^2 + c_3^2 + 2ac_3c_4  \\
\end{cases}
is equivalent to
\begin{cases}
a=c_{1}\dfrac{c_{3}^{3}-x^{3}}{c_{2}x^{3}}\\[2ex]
b=x ,
\end{cases}
where $x$ is a solution of the following septic equation I've obtained in SWP:
\begin{eqnarray*}
0 &=&c_{2}^{2}x^{7}+c_{3}c_{2}^{2}x^{6}+\left(
-c_{1}^{2}+2c_{1}c_{3}c_{4}c_{2}\right) x^{5}+\left(
2c_{3}^{2}c_{1}c_{4}c_{2}-c_{3}c_{1}^{2}\right) x^{4} \\[2ex]
&&+\left( 2c_{3}^{3}c_{1}c_{4}c_{2}-c_{1}^{2}c_{3}^{2}\right)
x^{3}+c_{1}^{2}c_{3}^{3}x^{2}+c_{1}^{2}c_{3}^{4}x+c_{1}^{2}c_{3}^{5}.
\end{eqnarray*}

I am stuck at this point.

The general septic equation cannot be solved algebraically.
Note: In the present form of the system $a$ should be different from $0$. So $b=c_{3},a=0$
is no longer a solution. 
