<s>[INST] Show that $n^3 > (n+1)^2$ for $n>2$ using mathematical induction Is the following a correct way of showing that $n^3 > (n+1)^2$ for $n>2$ using mathematical induction? Thank you in advance!
$P(n): n^3>(n+1)^2$
$P(3): 3^3>(3+1)^2$
$27 > 16$ (correct)
$P(k): k^3>(k+1)^2$
Assume $P(k)$ is true.
$P(k)$ $\implies$ $P(k+1)$?
$k^3+3k^2+3k+1>(k+1)^2+3k^2+3k+1$
$(k+1)^3>(k+1)^2+3k^2+3k+1$
$(3k-2)(k+1)>0$ $\forall$ $k\in N$
$\implies 3k^2+k-2>0$
$\implies 4k^2+5k+2>k^2+4k+4$
$\implies (k^2+2k+1)+3k^2+3k+1>(k+2)^2$
$\implies (k+1)^2+3k^2+3k+1>(k+2)^2$
Hence $P(k) \implies P(k+1)$
Hence by the principal of mathematical induction $P(n)$ is true $\forall n>2$  
 [/INST] That looks fine, but here is a simpler way to do the inductive step: assuming $k\ge3$ and $k^3>(k+1)^2$, we have
$$\eqalign{(k+1)^3
  &=k^3+3k^2+3k+1\cr
  &>(k+1)^2+3+2k+0\cr
  &=k^2+4k+4\cr
  &=(k+2)^2\ .\cr}$$
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