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Not sure where to go with this

$2^k > k^3$ for $k > 9$

$2^(k+1) > (k+1)^3$

$2^(k+1) = 2^k \cdot 2$

$2^k \cdot 2 > k^3 \cdot 2$ (by inductive hypothesis)

$2^k \cdot 2 > 2k^3$

$2k^3 > k^3 + 3k^2 + 3k + 1$

$2k^3 > (k+1)^3$

I know the final inequality is true, since I graphed it, but I was wondering if there was a clearer way to show the thought process algebraically.

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You have $2^{k+1}=2\cdot2^k>2k^3$, so it suffices to show that $2k^3\ge(k+1)^3$, or, equivalently, that $k^3\ge 3k^2+3k+1$ for $k>9$. And

$$3k^2+3k+1<3\left(k^2+k+1\right)\le3\left(3k^2\right)=9k^2<k^3$$

for $k>9$.

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You want to show that $2^k > k^3$ implies $2^{k+1} > (k+1)^3$ for $k > 9$.

Let's try proof by contradiction.

Suppose $2^k > k^3$ and $2^{k+1} \le (k+1)^3$. Then $(k+1)^3 \ge 2^{k+1} = 2\ 2^k > 2\ k^3 $ or $k^3 < 3k^2+3k+1$ which, by a non-coincidence, is the same as Brian M. Scott's inequality, and can be shown false the same way for $k > 9$.

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