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?

 A: Define $$R_n=(a_1^n+a_2^n+a_3^n -b_1^n-b_2^n-b_3^n)/n$$
I found and proved that 
$$R_1S+R_3^3-2R_2R_3R_4+R_2^2R_5=\prod_{i=1}^3\prod_{j=1}^3\big(a_i-b_j\big)$$
where $S$ is a polynomial containing 9 items.
If $R_1=R_2=0$, then $$R_3^3=\prod_{i=1}^3\prod_{j=1}^3\big(a_i-b_j\big)$$
Example: $2^n+13^n+21^n=6^n+7^n+23^n$, $(n=1,2)$
If $R_1=R_3=0$, then $$R_2^2R_5=\prod_{i=1}^3\prod_{j=1}^3\big(a_i-b_j\big)$$
Example: $2^n+10^n+12^n=3^n+8^n+13^n$, $(n=1,3)$
If $R_1=R_4=0$, then $$R_3^3+R_2^2R_5=\prod_{i=1}^3\prod_{j=1}^3\big(a_i-b_j\big)$$
Example: $3^n+25^n+38^n=7^n+20^n+39^n$, $(n=1,4)$
If $R_1=R_5=0$, then $$R_3^3-2R_2R_3R_4=\prod_{i=1}^3\prod_{j=1}^3\big(a_i-b_j\big)$$
Example: $3^n+54^n+62^n=24^n+28^n+67^n$, $(n=1,5)$
Let $a_i=x_i^2,b_i=y_i^2$, $(i=1,2,3)$, the system $(1)$ satisfies $R_1=R_3=0$, then we can prove equation $(2)$.
For more identities of the similar form, please refer to my website on Algebraic Identities
For more examples of numerical solutions, please refer to my website on Equal Sums of Like Powers
A: 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.
