For a certain base $b$, the product $(12_b)(15_b)(16_b)$ is equal to $3146_b$. Let $s = 12_b + 15_b + 16_b$. What is $s$ in base $b$? I have worked out the above problem in the following way:
$(12_b)(15_b)(16_b)=(b+2)(b+5)(b+6)=3146_b=3b^3+b^2+4b+6$
$=>b^3+13b^2+52b+60=3b^3+b^2+4b+6$
$=>-2b^3+12b^2+48b+54=0$
$=>-b^3+6b^2+24b+27=0$
$=>6b^2+24b+27=b^3$
=>$2b^2+8b+9=b^3$
We need to find $s=3b+13$. I am stuck there. Any hint on how to proceed would be appreciated.

Well done, until $6b^2+24b+27=b^3 \implies 2b^2+8b+9=b^3$. This step is wrong.
From $6b^2+24b+27=b^3$ we get that $b$ is a multiple of $3$. Since $6$ appears in $16_b$, we have $b>6$.
By inspection, $b=9$ is a root of $6b^2+24b+27=b^3$.
