Why is $2n^3 > (n+1)^3$? I am doing an induction proof and one of the intermediate steps is proving that $2n^3 > (n+1)^3$ when $n > 9$. Can someone show a short proof of why $2n^3 > (n+1)^3$ when $n>9$? I cannot see it.
Thank you
 A: $2n^3 > (n+1)^3 \iff n^3 > 3n^2 + 3n + 1$. Assume that $n \ge 7\implies n^3 \ge 7n^2 = 3n^2 + 3n^2 + n^2 \ge 3n^2 + 3n + 1$. Thus you're done. 
A: $$
2n^3>(n+1)^3\Longleftrightarrow\left(\frac{2^{1/3}n}{n+1}\right)^3>1\Longleftrightarrow\frac{2^{1/3}n}{n+1}>1\Longleftrightarrow n>\frac{1}{2^{1/3}-1}\approx 3.847.
$$
A: Note that :$(2^{1/3}-1)n=0.2599....>1$ when $n>3$ 
so $$n+1<n+(2^{1/3}-1)n= 2^{1/3}n\\ when\ n>3$$
We inequality by raising the third power.
A: Another approach, one that takes a little calculation but that also works: $2n^3\gt (n+1)^3$ iff $\dfrac{(n+1)^3}{n^3}\lt 2$ iff $\dfrac{(n+1)}{n}\lt\sqrt[3]{2}$ iff $1+\frac1n\lt\sqrt[3]{2}$ iff $\frac1n\lt\sqrt[3]{2}-1$ iff $n\gt\dfrac{1}{\sqrt[3]{2}-1}$.  Now it's just calculation; you can show that $\sqrt[3]{2}\gt \frac54$ (since $2=\frac{128}{64}\gt\frac{5^3}{4^3}=\frac{125}{64}$), so the equation holds for all $n\gt 4$.
A: $2n^3 > (n+1)^3 \iff n^3 > 3n^2 + 3n + 1$. None of the inequalities holds for $n\leq 3$.
But for $n\geq5$,
$n^3=n \times n^2> 4 \times n^2= 3n^2+n^2>3n^2+3n+1$ (Because, $n^2>3n+1$ for $n>3$).
So, $2n^3 > (n+1)^3$ holds for $n\geq 5$.
Moreover, it can be verify that it holds for $n=4$.
Hence, $2n^3 > (n+1)^3$ holds for $n\geq 4$.
