# Math Help - Second Question about Reversing Equations

1. ## Second Question about Reversing Equations

Here another similar question but more complicated than former:

$
\begin{cases}
& u=a^3+3b^2d+3bc^2+3ce^2+3d^2e+6abe+6acd \\
& v=b^3+3a^2d+3ae^2+3cd^2+3c^2e+6abc+6bde \\
& y=d^3+3a^2e+3ac^2+3b^2c+3be^2+6abd+6cde \\
\end{cases}$

I would interesting how many solutions when all variables are real numbers. And difference with when all variables are complex numbers.

2. Originally Posted by haedious
Here another similar question but more complicated than former:

$
\begin{cases}
& u=a^3+3b^2d+3bc^2+3ce^2+3d^2e+6abe+6acd \\
& v=b^3+3a^2d+3ae^2+3cd^2+3c^2e+6abc+6bde \\
& y=d^3+3a^2e+3ac^2+3b^2c+3be^2+6abd+6cde \\
\end{cases}$

I would interesting how many solutions when all variables are real numbers. And difference with when all variables are complex numbers.
The solution in the complex case will be similar to the previous problem. Let $\omega = e^{2\pi i/5}$, and for k=0,1,2,3,4 let $\lambda_k = u+\omega^{-k}v + \omega^{-2k}x + \omega^{-3k}y + \omega^{-4k}z.$

Then $(a + \omega^{k}b + \omega^{2k}c + \omega^{3k}d + \omega^{4k}e)^3 = \lambda_k$, and it follows that

$\displaystyle a = \tfrac15\sum_{k=0}^5\lambda_k^{1/3},\quad b = \tfrac15\sum_{k=0}^5\omega^k\lambda_k^{1/3},\quad c = \tfrac15\sum_{k=0}^5\omega^{2k}\lambda_k^{1/3},\quad d = \tfrac15\sum_{k=0}^5\omega^{3k}\lambda_k^{1/3},\quad e = \tfrac15\sum_{k=0}^5\omega^{4k}\lambda_k^{1/3}.$

Each $\lambda_k$ has three complex cube roots, and there are five of them, giving a total of $3^5 = 243$ complex solutions. The number of real solutions is much more problematic. It's not clear to me whether there will be any at all.

3. Originally Posted by Opalg
The solution in the complex case will be similar to the previous problem. Let $\omega = e^{2\pi i/5}$, and for k=0,1,2,3,4 let $\lambda_k = u+\omega^{-k}v + \omega^{-2k}x + \omega^{-3k}y + \omega^{-4k}z.$

Then $(a + \omega^{k}b + \omega^{2k}c + \omega^{3k}d + \omega^{4k}e)^3 = \lambda_k$, and it follows that

$\displaystyle a = \tfrac15\sum_{k=0}^5\lambda_k^{1/3},\quad b = \tfrac15\sum_{k=0}^5\omega^k\lambda_k^{1/3},\quad c = \tfrac15\sum_{k=0}^5\omega^{2k}\lambda_k^{1/3},\quad d = \tfrac15\sum_{k=0}^5\omega^{3k}\lambda_k^{1/3},\quad e = \tfrac15\sum_{k=0}^5\omega^{4k}\lambda_k^{1/3}.$

Each $\lambda_k$ has three complex cube roots, and there are five of them, giving a total of $3^5 = 243$ complex solutions. The number of real solutions is much more problematic. It's not clear to me whether there will be any at all.
Thanks again!
About the real solutions I'm thinking if $n^{1/4}$ will outcome 2 real numbers: $\sqrt{n},-\sqrt{n}$ and 2 other complexs $\sqrt{n}i,-\sqrt{n}i$. But I agree it could be more problematic for sort of this questions.