Solution
= Solution
\b[False.] Put $r_n=3^n/n^3$ for $n\geq1$. For $n\geq4$,
$$
\frac{r_{n+1}}{r_n}=3\left(\frac n{n+1}\right)^3
\geq3\left(\frac45\right)^3=\frac{192}{125}>1.
$$
Thus $r_n\geq r_4(192/125)^{n-4}\to\infty$. No fixed constant can bound $3^n/n^3$ eventually, so $3^n\ne O(n^3)$. This is an explicit comparison of <exponential growth> with <polynomial growth>.