How to prove this nice $10$th power identity for $x_1^6+x_2^6+x_3^6 =y_1^6+y_2^6+y_3^6$? Ramanujan's 6-10-8 Identity turns out to depend on a special case of,
$$u_1^k+u_2^k+u_3^k =v_1^k+v_2^k+v_3^k$$
simultaneously valid for $k=2,4$. I was investigating if the next system $k=2,6$,
$$x_1^2+x_2^2+x_3^2 =y_1^2+y_2^2+y_3^2\\x_1^6+x_2^6+x_3^6 =y_1^6+y_2^6+y_3^6\tag1$$
would have something similar. I observed empirically that,
$$\left(\sum_{i=1}^3\big(x_i^{10}-y_i^{10}\big)\right)\left(\sum_{i=1}^3\big(x_i^{4}-y_i^{4}\big)\right)^2=20\prod_{i=1}^3\prod_{j=1}^3\big(x_i^2-y_j^2\big)\tag2$$
Example:
$$10^k+15^k+23^k = 3^k+19^k+22^k$$
yields,
$$\small \text{LHS}= \big(10^{10} + 15^{10} + 23^{10} - 3^{10} - 19^{10} - 22^{10}\big)\big(10^4 + 15^4 + 23^4 - 3^4 - 19^4 - 22^4\big)^2$$
$$\small \text{RHS}=20(10^2 - 3^2)(10^2 - 19^2)(10^2 - 22^2)(15^2 - 3^2)(15^2 - 19^2)(15^2 - 22^2)(23^2 - 3^2)(23^2 - 19^2)(23^2 - 22^2)$$
$$\small\text{LHS}=\text{RHS}=37739520^2\times3830610$$
I've also tested it with more general parametric solutions and it works just fine.

Q: But how do we prove $(2)$ rigorously?


You want to prove equation $(2).$ Using equation $(1)$ with $k=2$ solve the quadratic for say $y_3$. Substitute this value of $y_3$ into  LHS-RHS  of equation $(1)$ with $k=6$ to get a homogeneous $6$th degree polynomial in the $5$ other variables. Substitute the value of $y_3$ into LHS-RHS of equation $(2)$ and verify that the result is a homogeneous $18$th degree polynomial divisible by the homogeneous $6$th degree polynomial.
