Prove $27195^8 - 10887^8 + 10152^8\equiv 0\pmod{26460}$ So far I've tried:
1) $26460 = 2^2 * 3^3 * 5 * 7^2$
2) $10152 = 2^3 * 3^3 * 47$
3) $27195 - 10887 = 16308 = 2^2 * 3^3 * 151$ (I know $a^8 - b^8\equiv0\pmod{a - b}$)
Therefore I conclude that $27195^8 - 10887^8 + 10152^8$ is divisible by $2^2 * 3^3$, as well is 1).
But what about $5 * 7^2$ part? 
 A: Let $a=27195, b=10887, c=10152$ and consider this:
$$
\matrix{
m       &a \bmod m      &b \bmod m      &c \bmod m      \\
2^2     &3      &3      &0      \\
3^3     &6      &6      &0      \\
5\hphantom{^1}       &0      &2      &2      \\
7^2     &0      &9      &9      \\
}
$$
Note how for each $m$, we have $b \equiv a$ or $b \equiv c \bmod m$, and the other is $0 \bmod m$.
Therefore, the terms in $27195^8 - 10887^8 + 10152^8$ reduce to $0 \bmod m$.
A: Hint $\ $  Exploit innate symmetry using the following observation
$$\begin{eqnarray}a\equiv c,&&\ b\equiv d\!\!\pmod{m}\\
a\equiv d,&&\ b\equiv c\!\!\pmod{k}\end{eqnarray}\, \Rightarrow\,\ e := a^n+b^n-c^n-d^n\equiv 0\!\! \pmod{{\rm lcm}(m,k))}$$
because $\ \ {\rm mod}\ m\!:\ a\equiv c\,\Rightarrow a^n\equiv c^n,\ \ b\equiv d\,\Rightarrow\ b^n\equiv d^n\,\Rightarrow\,e\equiv 0\, $ by Congruence Power Rule. 
Similarly $\, {\rm mod}\ k\!:\,\ e\equiv 0.\ $ Thus $\ m,k\mid e\,\Rightarrow\,{\rm lcm}(m,k)\mid e$.
Remark $\ $ More generally, bringing to the fore the innate symmetry yields:
Assume  $ \{a,b\}\equiv \{c,d\}\ {\rm mod}\ m,k\ $ and $\,f(x,y)\,$ is symmetric $\,f(x,y)=f(y,x).$
Then $\,\ \ f(a,b)\equiv f(c,d)\ {\rm mod}\ m,k\,$ so also mod $\,{\rm lcm}(m,k).$ 
OP has  $\, f(x,y) = x^n + y^n.$
Remark $\ $ See here and here for some explicit examples. 
