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Would you please check if my answer is correct and confirm that it is proof by induction? Thank you.

The proof is by induction.

Base Case: when $n=1$:

$$\begin{eqnarray}1^{3}&>&1^{2}-6(1)+4\\ 1&>&-1\end{eqnarray} $$

Thus, the base case is proven.

Inductive Step: suppose $n^{3}>n^{2}-6n+4$ and $n\ge2 , n\in\mathbb{N}$ .

$$ \begin{eqnarray} n^{3}+(3n^{2}+3n+1) & > & n^{2}-6n+4+(3n^{2}+3n+1) \\ n^{3}+3n^{2}+3n+1 & > & 4n^{2}-3n+5 \\ n^{3}+3n^{2}+3n+1 & > & n^{2}+2n+1-6n-6+4+3n^{2}+n+6 \\ (n+1)^{3} & > & (n+1)^{2}-6(n+1)+4+3n^{2}+n+6 \\ & > & (n+1)^{2}-6(n+1)+4 \\ \end{eqnarray} $$ since $n\ge2 , n\in\mathbb{N} $.

This proves the inductive step, so the result follows by induction.

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    Overall correct. Style is high school two column, most lines, starting with $1^3>1^2-6(1)+4$. Should change, unless you are in high school. Small almost typo, want $n\ge 1$. Else in principle have excluded step from $1$ to $2$. I prefer going from $n=k$ to $n=k+1$.2012-05-02

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