Prove $n^3>2n-2$ for all $n∈N $
step 1: claim (1) is 1 > 0
LHS > RHS
step 2: assume claim (k) is true, that is $k^3>2k-2$
Prove claim (k + 1)
$(k+1)^3>2(k+1)-2$
LHS = $(k+1)^3$
= $k^3+3k^2+3k+1$
$>(2k-2)+3k^2+3k+1$
Then I am struck, I am not sure how to get to $2(k+1)-2$
Hint: $$3k^2+3k = 3k(k+1)\ge 6 >1.$$