<s>[INST] Proof technique regarding divisibility by $n^3$ I read in a note that-

Let, $z \geq3, n$ is even, and 
$$(18n^2 + 1)^x + (7n^2 − 1)^y = (5n)^z$$ By taking modulo $n^3$
  implies that
$$1 + 18n^2x − 1 + 7n^2y ≡ 0 \pmod{ n^{3}}$$

I couldn't get how $1 + 18n^2x − 1 + 7n^2y ≡ 0 \pmod{ n^{3}}$ is derived,  how it can be proved?
 [/INST] This is just use of the binomial theorem: $$(a+b)^x={x\choose 0}a^x+{x\choose 1}a^{x-1}b+{x\choose 2}a^{x-2}b^2...+ {x\choose x-1}b^{x-1}a +{x\choose x}b^x$$
where $a=1$ and $b = 18n^2$
$$(1+18n^2)^x=1+ x18n^2+{x(x-1)\over 2}18^2n^4+...+18^xn^{2x}$$
$$=1+ x18n^2+n^3\Big({x(x-1)\over 2}18^2n+...+18^xn^{2x-3}\Big)$$
$$ \equiv 1+x18n^2 \pmod{n^3}$$
and similary for the second bracket.
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