Question: Let $f$ be an entire function such that $|f(z)|\leq1+2|z|^{10/3}$ for all z. Prove that $f$ is a cubic polynomial
Thoughts so far: Using a corollary of Liouville's theorem, we know that we want to show that $|f(z)|\leq a+b|z|^3$ and $|f(z)|\geq a+b|z|^3$ for some constants a and b. We know that within the unit circle $|f(z)|\leq 1+2|z|^{10/3} < 1+2|z|^3$ which gives us an upper bound, while outside of the unit circle we know that $-|f(z)|\geq -1-2|z|^{10/3} \implies |f(x)| \geq |-1-2|z|^{10/3}| = |2|z|^{10/3}--1|$ (by triangle inequality) $\geq 2|z|^{10/3}-1 > 2|z|^3-1$, which provides an lower bound of three, which by the corollary of Liouville's theorem implies that f(z) must be cubic. However, this proof makes me quesy because I feel that the upper and lower limits were chosen arbitrarily and could be any such function with a power less than $\frac{10}{3}%$, which makes me feel rather frustrated. Furthermore, this also leads me to believe that this is not a constructive line of thought for this problem.
Thank you in advance for any help that you may provide.