<s>[INST] 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.
 [/INST] Step: $m+1$:
$m^3= (m+1-1)^3= $
$(m+1)^3 -3(m+1)^2+3(m+1) -1\lt 2^m$
(Hypothesis).
$(m+1)^3 \lt $
$2^m +1-3(m+1) +3(m+1)^2<$
$2^m +1 + 3(m+1)^2 <2^m +2^m =2^{m+1};$
Used : 
$1+3(m+1)^2 =$
$3m^2+6m +4 \lt $
$6m^2+4 < 2^m$ , for $m \ge 10.$
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