# 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)

-
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

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.

-

$a=(\frac{c_1}{c_2})((\frac{c_3}{b})^3-1)$

$a^2=(\frac{c_1}{c_2})^2((\frac{c_3}{b})^3-1)^2$

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

-

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 :)

-
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