Help!

As follows:

The sum of 3 real numbers is known to be zero.
If the sum of their cubes is $\pi^e$, what is their product equal to?

$x+y+z = 0 \\ x^3+y^3+z^3 = \pi^e$

1.)

$\begin{array}{rcll} 0&=&(x+y+z)^3 \\ &=& x^3+y^3+z^3 +6xyz+3\left( x^2(y+z)+y^2(x+z)+z^2(x+y)\right) \end{array}$

2.)

$\begin{array}{rcll} 0&=&(x+y+z)(x^2+y^2+z^2) \\ &=& x^3+y^3+z^3 + x^2(y+z)+y^2(x+z)+z^2(x+y) \\ x^2(y+z)+y^2(x+z)+z^2(x+y) &=& -(x^3+y^3+z^3) \\ \end{array}$

$\begin{array}{rcll} x^3+y^3+z^3 +6xyz+3\left( x^2(y+z)+y^2(x+z)+z^2(x+y)\right) &=& 0 \\ x^3+y^3+z^3 +6xyz+3\left( -(x^3+y^3+z^3) \right) &=& 0 \\ -2(x^3+y^3+z^3) +6xyz &=& 0 \\ 6xyz &=& 2(x^3+y^3+z^3) \\ xyz &=& \dfrac{1}{3}(x^3+y^3+z^3) \\ \mathbf{xyz} &=& \mathbf {\dfrac{ \pi^e}{3}} \\ \end{array}$