About Big O properties
If $f(x) = O(x^n)$ then $f(x)^m = O(x^{nm})$. Indeed, $f(x) = O(x^n)$ means that $f(x) \leq Cx^n$ for some $C&gt;0$. Therefore $f(x)^m \leq C^mx^{nm}$, and so $f(x)^m = O(x^{nm})$.

On the other hand, it is not true that $(g(x) + h(x))^m = g(x)^m + O(h(x)^m)$. For example, suppose that $g(x) = 2$ and $h(x) = 1$. Then it is not the case that $3^m = 2^m + O(1)$.

What is true is that $(g(x) + h(x))^m = O(g(x)^m + h(x)^m)$. Indeed, if $g(x) \leq h(x)$ then $$(g(x) + h(x))^m \leq (2h(x))^m = 2^m h(x)^m = O(h(x)^m) = O(g(x)^m + h(x)^m), $$
and similarly in the other case.

(Throughout this answer, $m \geq 1$ is a constant. The first part holds for any constant $m \geq 0$.)