Proof technique regarding divisibility by $n^3$

Question

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?

Answer 1

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.