let (n^3) + 2 be f(n)
since proof is for n > 9
let n be 10,
f(10) = 1002 < 2^10 = 1024
when n = g (arbitiary value), where g>9
f(g) = g^3 + 2 < 2^g -------- result (1) -- you have to hold this to be true
now take n = g+1, where g>9
f(g+1) = (g+1)^3 + 2 = g^3 + 3g^2 + 3g + 3
from result (1),
2g^3 + 2 < 2*2^g = 2^(g+1) --- result (2)
taking the difference between f(g+1) and result (2),
g^3 - 3g^2 - 3g + 1 has to be greater or equal to 0 for f(g+1) < result (2) [basic algebra rearrangement]
this is true when g > 9
as g^3 - 3g^2 -3g + 1 = (g+1) [(g-2)^2 -3]
if the above is true, and statement is true when n = 10 > 9
then by mathematical induction, the statement is true
btw i did this quickly, and something's not right because if you follow the steps, it seems to suggest that the statement is true for values smaller than n>9 (which isn't true, if you check the results), maybe someone at the jc level who's still familiar with mathematical induction can help.