Let $ad=bc$. Then Ramanujan's 6-10-8 Identity is the bizarre,
$$\small 64[(a+b+c)^6+(b+c+d)^6-(c+d+a)^6-(d+a+b)^6+(a-d)^6-(b-c)^6]\\ \small [(a+b+c)^{10}+(b+c+d)^{10}-(c+d+a)^{10}-(d+a+b)^{10}+(a-d)^{10}-(b-c)^{10}] =45[(a+b+c)^8+(b+c+d)^8-(c+d+a)^8-(d+a+b)^8+(a-d)^8-(b-c)^8]^2$$
It can be somewhat de-mystified. For generic $a,b,c,d,e,f$, define,
$$F_k = a^k+b^k+c^k-(d^k+e^k+f^k)$$
If $F_2 = F_4 = 0$ and $a+b+c = d+e+f = 0$, then the 6-10-8 is,
$$64F_6 F_{10} = 45F_8^2$$
and Hirschhorn gave,
$$25F_3 F_7 = 21F_5^2$$
I found this has an odd power counterpart. Define,
$$G_k = a^k+b^k+c^k+d^k-(e^k+f^k+g^k+h^k)$$
If $G_1 = G_3 = G_5 = 0$ and $a+b+c+d = e+f+g+h = 0$, then it can be observed that,
$$7G_4G_9 = 12 G_6G_7$$
One can test it with a random example {$a,b,c,d$} = {$21,\, 9, -13, -17$} and {$e,f,g,h$} = {$23,\, 1, -3, -21$} but it will work in general.
Question: The next step seems obvious: to use more terms and higher powers. Anybody knows how to find if there are higher power versions?
I found simplified equivalent expressions by changing $F_k$ to $P_k$ $($where $P_k=F_k/k)$.
Define $$P_k=(a_1^k+a_2^k+a_3^k-b_1^k-b_2^k-b_3^k)/k$$ If $P_1=P_2=P_4=0$, then the Ramanujan's 6-10-8 Identity is, $$4P_6P_{10}=3P_8^2$$ and the Hirschhorn's 3-7-5 Identity is, $$P_3P_7=P_5^2$$ and the identities by Chamberland are, $$P_3P_8=2P_5P_6$$ $$P_3P_{10}=3P_6P_7$$
I found more interesting results, $$\frac{P_5}{P_3}=\frac{P_7}{P_5}=\frac{P_8}{2P_6}=\frac{2P_{10}}{3P_8}=\sqrt{\frac{P_7}{P_3}}=\sqrt{\frac{P_{10}}{3P_6}}$$ $$\frac{P_8}{2P_5}=\frac{P_6}{P_3}=\frac{P_{10}}{3P_7}$$ $$\frac{3P_5^2}{P_{10}}=\frac{P_3^2}{P_6}=\frac{-P_{-1}^2}{P_{-2}}$$
For higher power, define, $$Q_k=(a_1^k+a_2^k+a_3^k+a_4^k-b_1^k-b_2^k-b_3^k-b_4^k)/k$$ If $Q_1=Q_2=Q_3=Q_5=0$, then the 4-9-6-7 identity can be simplified as, $$Q_4Q_9=2Q_6Q_7$$
I found the next higher power version. Define, $$R_k=(a_1^k+a_2^k+a_3^k+a_4^k+a_5^k-b_1^k-b_2^k-b_3^k-b_4^k-b_5^k)/k$$ If $R_1=R_2=R_3=R_4=R_6=0$, then, $$R_5(R_8^2+2R_7R_9-R_5R_{11})=R_7^3$$ Example $$a_1=-23,\ a_2=-10,\ a_3=-5,\ a_4=14,\ a_5=24$$ $$b_1=-21,\ b_2=-16,\ b_3=2,\ b_4=10,\ b_5=25$$ For more numeral examples, see my site on Equal Sums of Like Powers.
For more similar identities, see my site on The Generalization of Ramanujan’s 6-10-8 identity
or refer to the following pictures.
Simplified equivalent expressions of Ramanujan’s identity found by Chen Shuwen
Higher power version of Ramanujan’s 6-10-8 identity found and proved by Chen Shuwen