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.
 A: Hint: Use the fact that $k\geq 10$ so $2^k \geq 3k^2+3k+1$
A: Hint:
Use this fact that for all $k\geq10$ we have $k^3\geq1+3k^2+3k$. You can prove the latter claim b the induction on $k$ very easily. Now $2k^3\geq(1+k)^3$ and then $$2^{k+1}\geq (k+1)^3$$
A: Let $P(k)$ be $(k\geq 10\land k^3\leq 2^k).$ Since you have verified that $P(10)$ is true, what you want to do is to show that $P(k)\implies P(k+1).$
We have $$P(k)\iff (k\ge 10 \land k^3\leq 2^k)\implies$$ $$\implies (...\;k\geq 10 \land (k+1)^3=(1+\frac {1}{k})^3\cdot k^3\leq$$ $$\leq  (1+\frac {1}{k})^3\cdot 2^k\leq$$ $$\leq 2\cdot 2^k=2^{k+1}\;...)\implies$$ $$\implies (k+1\geq 10 \land (k+1)^3\leq 2^{k+1})\iff P(k+1).$$ 
This implicitly uses the fact that $k\geq 10\implies (1+\frac {1}{k})^3\leq (1+\frac {1}{10})^3\leq 2.$
The technique is to see how the information in $P(k)$ influences the assertion $P(k+1)$. The idea is that if $\frac {(k+1)^3}{k^3}\leq \frac {2^{k+1}}{2^k}$ then $P(k)\implies P(k+1).$
A: 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 $m \ge 10.$
