<s>[INST] 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)
 [/INST] $a=(\frac{c_1}{c_2})((\frac{c_3}{b})^3-1)$
$a^2=(\frac{c_1}{c_2})^2((\frac{c_3}{b})^3-1)^2$
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
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