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\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)

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Is it $$\begin{cases} a=\dfrac{c_1}{c_2} \left( (\dfrac{c_3}{b})^3 - 1 \right) \\[2ex] b^2 = a^2 + c_3^2 + 2ac_3c_4 \\ \end{cases}$$ I am stuck at this point. Not sure on how to move forward. One solution is $a=0,b=c_3$, but the general solution involves solving a 7th. degree equation! – Américo Tavares Mar 19 '14 at 18:09
So .... Any Idea on how to proceed with that? :) – Gowtham Mar 19 '14 at 18:23
What do the subscripts mean? – Daniel Pendergast Mar 19 '14 at 18:26
They are random constants – Gowtham Mar 19 '14 at 18:27
In my comment above "I am stuck at this point. Not sure on how to move forward." are two sentences of yours that I copied unintentionally. – Américo Tavares Mar 19 '14 at 18:30
up vote 1 down vote accepted

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.

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substitute $a^2$ and $a$ in the second equation






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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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 :)

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Yeah ... But then that turns out to be a 4th degree equation which is not that easy to be solved for ... Or Am I missing something? :) – Gowtham Mar 19 '14 at 17:59

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