Quite simply turned out to solve this Diophantine equation, when he made the assumption that the solutions of these equations symmetric.
So given this equation:
$$x^3+y^3+z^3+q^3+k^3=xyz+xyq+xyk+xzq+xzk+xqk+yzq+yzk+yqk+zqk$$
And symmetric solution is quite simple written.
$$x=25s^2+10ps+5p^2$$
$$y=10s^2+10ps+4p^2$$
$$z=20s^2+10ps+2p^2$$
$$q=5s^2+3p^2$$
$$k=15s^2+p^2$$
$s,p$ - integers of any sign.
The question is. This equation only symmetric solution? If not, what should be the idea for the solution of this equation?
In General, for any equation like this:
$$x^3+y^3+z^3+q^3+d^3+k^3=xyz+xyq+xyd+xyk+xzq+xzd+xzk+xqd+xqk+$$
$$+xdk+yzq+yzd+yzk+yqd+yqk+ydk+zqd+zqk+zdk+qdk$$
Symmetric solution can be written:
$$x=(3a^2+5b^2)p^2+5(9a^2+15b^2)s^2$$
$$y=80abps+200(a^2+5b^2)s^2$$
$$z=(a^2+15b^2)(p^2+15s^2)$$
$$q=5(a^2\pm2ab+5b^2)p^2+10(\pm5a^2+2ab\pm25b^2)ps+25(5a^2\mp6ab+25b^2)s^2$$
$$d=2(2a^2\pm5ab+5b^2)p^2+50(\pm{a^2}+2ab\pm5b^2)ps+10(31a^2\mp15ab+140b^2)s^2$$
$$k=2(a^2\pm5ab+10b^2)p^2+50(\pm{a^2}+2ab\pm5b^2)ps+10(28a^2\mp15ab+155b^2)s^2$$
$a,b,p,s$ - integers, any sign.
This is not clear. Quite limited symmetric solutions or need to look for others? And what way of calculation for this would be most effective?