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)
 A: 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. 
A: $a=(\frac{c_1}{c_2})((\frac{c_3}{b})^3-1)$
$a^2=(\frac{c_1}{c_2})^2((\frac{c_3}{b})^3-1)^2$
substitute $a^2$ and $a$ in the second equation
$b^2=(\frac{c_1}{c_2})^2((\frac{c_3}{b})^3-1)^2+c_3^2+2c_3c_4(\frac{c_1}{c_2})((\frac{c_3}{b})^3-1)$
$b^2c_2^2=c_1^2(c_3^3-b^3)^2+c_3^2c_2^2b^3+2c_3c_4c_1c_2(c_3^3-b^3)$
$0=c_1^2c_3^6-2c_1^2c_3^3b^3+b^6c_1^2+c_3^2c_2^2b^3+2c_3c_4c_1c_2c_3^3-2c_3c_4c_1c_2b^3-b^2c_2^2$
$0=b^6c_1^2+b^3(c_3^2c_2^2-2c_1^2c_3^3-2c_3c_4c_1c_2)-b^2c_2^2+c_1^2c_3^6+2c_3c_4c_1c_2c_3^3$
$b^3c_3^2c_2^2-2c_1^2c_3^3b^3-2c_3c_4c_1c_2b^3=b^2c_2^2-b^6c_1^2-c_1^2c_3^6-2c_3c_4c_1c_2c_3^3$
here we see that $b=c_3$ will make the expresion valid(LHS=RHS),the possibility of other roots is not obvious,and is questioneable
A: sub your equational value for a into your second line (b squared etc) 
ie sub c1c2((c3b)3−1) into b2=a2+c23+2(a)(c3)(c4)
Hope this helps you :)
