Prove by mathematical induction that $3^n>2n^3$ I'm having trouble with this question:
"Prove by mathematical induction that for all integers $n\ge 6$, $3^n>2n^3$".
I got to $P(k)=2k^3<3^k$ and $P(k+1)=2(k+1)^3<3^{k+1}=2k^3+6k^2+6k+2<3^k*3$,
but I dont know how I can get $P(k+1)$ from $P(K)$...
Thanks

$P(k)$ is the statement $ 2n^3<3^n $.  Do not write "$P(k)=.....$"; $P(k)$ is not a mathematical value.
If we assume that then we have $2n^3 < 3^n$
So $2(n+1)^3 = 2n^3 + 6n^2 + 6n + 1$  And we have $2n^3 < 3^n$ so
$2(n+1)^3 =2n^3 + 6n^2 + 6n + 1< 3^n + 6n^2 + 6n + 1$
And $n\ge 6$ so $6n^2 \le n*n^2 =n^3$ and $6n+1 < 6n+n < 6n*n=6n^2 < n*n^3 < n^3$.
So 
$2(n+1)^3 =2n^3 + 6n^2 + 6n + 1< 3^n + (6n^2) + (6n + 1)$
$< 3^n + n^3 + n^3 = 3^n + 2n^3 \le 3^n + 3^n < 3^n + 3^n + 3^n$
$< 3*3^n = 3^{n+1}$.
So $2(n+1)^3 < 3^{n+1}$ and so the statement $P(k+1)$ is true.
