<s>[INST] 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.
 [/INST] We have $$(b+2)(b+5)(b+6)=b^3+13b^2+52b+60=3b^3+b^2+4b+6$$
It follows the equation $$b^3-6b^2-24b-27=(b-9)(b^2+3b+3)=0$$
Hence $b=9$ and $$11+14+15=\color{red}{40}$$
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