The number in question when evaluated turns out to be\[ \frac{b^{3n}-b^{2n+1}+7b^{2n}+b^{n+1}-7b^n-1}{3(b-1)}=\frac{(b^n-1)(b^{2n}+b^n(8-b)+1)}{3(b-1)}\]Let the sequence of these numbers be $x_n^3$. We shall show that $b-1$ is a power of $3$. Suppose that this is not the case. Then, $\exists p \in \mathbb{P}$ such that $p \neq 3$ and $p|b-1$. Note that $b^{2n}+b^n(8-b)+1 \equiv 9 \ncong 0 \pmod{p}$. Suppose $p \neq 2$. Then, equating the powers of $p$ on either side using LTE, we get $3v_p(x_n) = v_p(n)$ for sufficiently large $n$. Choosing large $n$ with $v_p(n)=1$ does the job. Otherwise, if $p=2$, again equate powers of $2$ on either side using LTE. We get $v_2(b+1)+v_2(n)-1=3v_3(x_n)$, for all sufficiently large $n$. But, we can easily choose large $n$ such that the LHS is not a multiple of $3$, contradiction. Therefore, $b-1=3^a$. Since $b \ge 9$, we have $3^a \ge 8 \implies a \ge 2$. Suppose that $a\ge 3$. Then, note that \[v_3(b^{2n}+b^n(8-b)+1) = v_3((b^{2n}-b^{n+1})+8(b^n-1)+9) = 2\]because $27|b-1 \implies 27|b^n-1$ and $27|b^{n+1}(b^{n-1}-1)=b^{2n}-b^{n+1}$. Again, looking at the powers of $3$ on either side, we have $3v_3(x_n)=2+v_3(n)$. Choosing large $n$ not divisible by $3$ gives a contradiction. Therefore, $a=2$, which means $b=10$. And in this case we do have \[ x_n^3 = \frac{(10^n-1)(10^{2n}-2.10^n+1)}{27}= \left( \frac{10^n-1}{3} \right)^3 \]

Observe that the quantity in question can be written as: $$\frac{b^{3n} - b^{2n+1} + 7 \cdot b^{2n} + b^{n+1} - 7 \cdot b^n - 1}{3(b-1)}.$$ Observe that as $n$ is large, the cube root of this quantity is close to: $$\frac{b^n - \frac{b-7}{3}}{(3(b-1))^{\frac13}}. \qquad (1)$$ It's easy to see from here that $3(b-1)$ must be a cube, and so $b = 9k^3 + 1$. Plugging this in, we know that for large enough $n$, the quantity in question must exactly be the cube of $(1).$ This easily implies that $b = 10$, we're done. $\square$