By induction showing that $m^3\le2^m$ for $m\ge10$ Show that $m^3\le2^m$ for $m\ge10$

My try:

Base case is true for $m=10$

Inductive Hypothesis: Assume $P(k)$ is true $\implies k^3\le2^k$

Now showing that $P(k+1)$ is true

$(k+1)^3\le2^{k+1}$

$\implies (k+1)^3\le k^3+1+3k^2+3k$

$\le 2^{k+1}+3k^2+3k+1($ from inductive hypothesis$)$

From here I could not proceed.

Can anyone explain how to proceed from here.
Hint: Use the fact that $k\geq 10$ so $2^k \geq 3k^2+3k+1$
General solution:

For $m\ge m_0$,
$$m^k\le ab^m\implies m^{k+1}=m^k\left(\frac{m+1}m\right)^k\le m^k\left(\frac{m_0+1}{m_0}\right)^k\le ab^m\left(\frac{m_0+1}{m_0}\right)^k\le ab^{m+1}$$

provided

$$\left(\frac{m_0+1}{m_0}\right)^k\le b.$$

The base case is $$m_0^k\le ab^{m_0}.$$



Here we have a specific instance with

$$k=3,m_0=10,a=1,b=2.$$

We do verify $1.331\le2$ and $1000\le1024$.



Note that for any $b&gt;1$, you can pick

$$m_0=\left\lceil\frac1{\sqrt[k]b-1}\right\rceil$$ and $$a=\frac{m_0^k}{b^{m_0}}$$ which proves that all powers grow slower than all growing exponentials.