Given rationals $R = a,b,c,d,e,f$. Define,
$$F_n = a^n+b^n+c^n-(d^n+e^n+f^n)$$
- If $F_\color{red}1=0$, is there a rational solution to $7F_3x^4+7F_5x^2+F_7 = 0$? Then for $k=1,2,8$,
$$\small(x + a)^k + (x + b)^k + (x + c)^k + (x - d)^k + (x - e)^k + (x - f)^k = \\ \small(x - a)^k + (x - b)^k + (x - c)^k + (x + d)^k + (x + e)^k + (x + f)^k\tag1$$
- If $F_2=0$, is there a rational solution to $70F_4x^4+28F_6x^2+F_8 = 0$? Then for $k=1,2,3,8$,
$$\small(x + a)^k + (x + b)^k + (x + c)^k + (x - a)^k + (x - b)^k + (x - c)^k = \\ \small(x + d)^k + (x + e)^k + (x + f)^k + (x - d)^k + (x - e)^k + (x - f)^k\tag2$$
- If $F_2=0$, is there a rational solution to $42F_4x^4+28F_6x^2+3F_8 = 0$? Then for $k=1,2,3,9$,
$$\small(x + a)^k + (x + b)^k + (x + c)^k + (x - a)^k + (x - b)^k + (x - c)^k = \\ \small(x + d)^k + (x + e)^k + (x + f)^k + (x - d)^k + (x - e)^k + (x - f)^k\tag3$$
For the third, one solution is $R = 9, 14, 19;\, 17, 18, 5$, and $x = 4$. In fact, using an elliptic curve, one can show there are an infinite more. So this is a settled question.
However, I'm having trouble with the first and second which at first glance seems to be the easier problem. For the first, the best I could find is $R = 5, 19, 19;\, 11, 11, 21$ and $x = \sqrt{13}$. Using some basic Mathematica code, I searched $0<R<24$ as it took too long to search higher. Maybe someone here knows of a more efficient way.
P.S. See also this related post.