Solution
= Solution
Treat $a,b$ as fixed positive <integers>. For $n\geq1$, $n^2+a\leq(1+a)n^2$. Raising this inequality to the positive power $b$ gives
$$
(n^2+a)^b\leq(1+a)^b n^{2b}.
$$
The factor $(1+a)^b$ is independent of $n$, as required by <Big O notation>. Hence
$$
\boxed{\text{True for fixed }a,b:\quad(n^2+a)^b=O(n^{2b}).}
$$
A bound uniform over varying $a,b$ is not what the notation here asserts.