Polynomial Complexity (Relative to the Size of the Input)
The statement meant to say that "if b is asymptotically smaller than n" i.e. $b=o(n) \space or \space b=O(n)$. 
If $b$ was asymptotically larger than $n$, i.e. $b=\omega (n) \space or \space b=\Omega(n)$, then the complexity would become, $O(b^3)$, i.e. not polynomial in $n$.

For example, lets consider two cases.


$b$ is asymptotically smaller than $n$, i.e. $b=O(n)$.
$$O((n+mb)^3) = O(n^3 + 3n^2mb + 3nm^2b^2 + m^3b^3)$$
$$ = max(O(n^3),O(3n^2mb),O(3nm^2b^2),O(m^3b^3))$$
We know that $b=O(n) \implies bn=O(n^2),\space bn^2=O(n^3) \space and \space b^2n=O(n^3). $
From this, we can conclude that, $m^3b^3=O(3nm^2b^2)$, $3nm^2b^2=O(3n^2mb)$, and $3n^2mb = O(n^3)$.
$$\implies O((n+mb)^3)=O(n^3)$$
$b$ is asymptotically larger than $n$, i.e. $b=\Omega(n) \space or \space n=O(b)$.
$$O((n+mb)^3) = O(n^3 + 3n^2mb + 3nm^2b^2 + m^3b^3)$$
$$ = max(O(n^3),O(3n^2mb),O(3nm^2b^2),O(m^3b^3))$$
We know that $n=O(b) \implies nb=O(b^2),\space nb^2=O(b^3) \space and \space n^2b=O(b^3). $
From this, we can conclude that, $n^3=O(3n^2mb)$, $3n^2mb=O(3nm^2b^2)$, and $3nm^2b^2 = O(b^3)$.
$$\implies O((n+mb)^3)=O(b^3)$$


In your approach, assuming $b$ to be numerically smaller than $n$ would lead to correct analysis. But what the statement means is that $b$ is asymptotically smaller than $n$. Like, $b=log(n) \space or \space b=\sqrt{n} \space or \space b=log(log(n))$