Is it true that $f(n) = c\cdot g(n) + O(g(n))$ implies $f(n) = O(g(n))$?
Firstly let me say, that in definition of big-O, for non negative case, there is inequality(not equality, typo I think):
$$O(g)=\{f: \exists C&gt;0, \exists N \in \mathbb{N}, \forall n&gt; N, f(n)\leqslant C g(n)\}$$
Now  $f(n) = c\cdot g(n) + O(g(n))$ means, that  $\exists \phi (n) \in O(g(n))$ such that $f(n) = c\cdot g(n) + \phi (n)$. From here we have
$$f(n) = c\cdot g(n) + \phi (n) \leqslant c\cdot g(n) + C\cdot g (n)=(c+C)g (n)$$
so $f \in O(g)$ and your claim is correct.
The general rule can be worded as follows:
If $f \in O(g)$, then $f+O(g)=O(g)$