For $k = 0,1,2,3,4,5,6,7,8$, we have the equality,
$$(-5)^k + (-119)^k + (-101)^k + (-215)^k + (-197)^k + 43^k + 157^k + 31^k + 217^k + 169^k\\ =\\ (-47)^k + (-161)^k + (-35)^k + (-221)^k + (-173)^k + 1^k + 115^k + 97^k + 211^k + 193^k$$
I was very surprised to see the following more general identity (I have google it, and not found it)： post
\begin{align*} &(m^2-mn-n^2)^k+(15m^2-25mn-21n^2)^k+(13m^2-71mn+7n^2)^k\\ &+(27m^2-95mn-13n^2)^k+(-11m^2-27mn-33n^2)^k+(17m^2-129mn+71n^2)^k\\ &+(3m^2-105mn+91n^2)^k+(29m^2-101mn+51n^2)^k+(-m^2-49mn+79n^2)^k\\ &+(-13m^2-27mn+59n^2)^k\\ &=(-m^2+3mn-13n^2)^k+(13m^2-21mn-33n^2)^k+(-13m^2-25mn+7n^2)^k\\ &+(17m^2-77mn-21n^2)^k+(29m^2-99mn-n^2)^k+(15m^2-125mn+59n^2)^k\\ &+(m^2-101mn+79n^2)^k+(3m^2-55mn+51n^2)^k+(-11m^2-31mn+71n^2)^k\\ &+(27m^2-99mn+91n^2)^k \end{align*}
where the example is just the particular case $m,n = 1,2$.
Maybe this identity is new result? Because we only know some background:
http://mathworld.wolfram.com/DiophantineEquation5thPowers.html
http://mathworld.wolfram.com/DiophantineEquation6thPowers.html
http://mathworld.wolfram.com/DiophantineEquation7thPowers.html
http://mathworld.wolfram.com/DiophantineEquation8thPowers.html
Can someone explain this identity's secret?
PS: this post author is Zipei Nie, see Zipei Nie
The identity used two theorems from equal sums of like powers and is hardly meaningless nor unmotivated. The $(k,10,10)$ equation above can be labelled as,
$$x_1^k+x_2^k+\dots+x_{10}^k = y_1^k+y_2^k+\dots+y_{10}^k$$
Notice that we have,
$$S_1 = x_1+y_6 = x_2+y_7 = x_3+y_8 = x_4+y_9 = x_5+y_{10}$$
$$S_2 = x_6+y_1 = x_7+y_2 = x_8+y_3 = x_9+y_4 = x_{10}+y_5$$
$$S_1=S_2=2(8m^2-63mn+29n^2)$$
In the study of the Prouhet-Tarry-Escott Problem, this is what is known as a symmetric solution and it is a clue how Zipie Nie found it. Assume the notation,
$$[a_1,a_2,\dots,a_m]^k = a_1^k + a_2^k +\dots +a_m^k$$
Theorem 1: (Tarry-Escott) If,
$$[a_1,a_2,\dots,a_m]^k = [b_1,b_2,\dots,b_m]^k,\;\; \text{for}\, k = 1,3,\dots, 2n-1$$
then for any constant $c$,
$$[c+a_1,\dots, c+a_m, c-b_1,\dots, c-b_m]^k = [c-a_1,\dots, c-a_m, c+b_1,\dots, c+b_m]^k\\ \text{for}\, k=1,2,3,\dots, 2n$$
This explains why sums of appropriate terms from either side of the equality is a constant. We can then find the "seed" identity of $a_i,b_i$ that Nie used as the $(k,5,5)$,
$$(-7 m^2 + 62 m n - 30 n^2)^k + (7 m^2 + 38 m n - 50 n^2)^k + (5 m^2 - 8 m n - 22 n^2)^k + (19 m^2 - 32 m n - 42 n^2)^k + (-19 m^2 + 36 m n - 62 n^2)^k\\=\\ (-9 m^2 + 66 m n - 42 n^2)^k + (5 m^2 + 42 m n - 62 n^2)^k + (-21 m^2 + 38 m n - 22 n^2)^k + (9 m^2 - 14 m n - 50 n^2)^k + (21 m^2 - 36 m n - 30 n^2)^k$$
for $k=1,3,5,7.$
In turn, this 7th degree multi-grade, for a suitable permutation of the $a_i,b_i$, belongs to a family by Gloden-Sinha,
Theorem 2: (Gloden-Sinha) If,
$$u_1^k+u_2^k+u_3^k = v_1^k+v_2^k+v_3^k,\;\;\text{for}\,k = 2,4\\ \text{where}\;\;u_1+u_2-u_3 = 2(v_1+v_2-v_3) \neq 0\tag1$$
or alternatively,
$$(a+4e)^k + (b+4e)^k + (a+b)^k = (c+2e)^k + (d+2e)^k + (c+d)^k,\;\; \text{for}\,k = 2,4$$
where $e\neq 0$ then,
$$(-a-b-e)^k + (-c-e)^k + (b+e)^k + (a+e)^k + (c+d+e)^k = \\(3e)^k + (a+b+3e)^k + (-a-3e)^k + (-b-3e)^k + (d+e)^k,\;\; \text{for}\, k = 1,3,5,7$$
So in a nutshell, what Nie essentially did (and is no trivial feat) was to find a solution to $(1)$ in terms of quadratic forms, of which several are known already.
P.S. For similar "mysteries", see also my MathOverflow answer regarding Ramanujan's 6-10-8 Identity which uses even higher powers.