To approach this, we start by writing out the definition of Θ with f(n) = 4n3 and g(n) = n3: Show that ∃ c1, c2, n0≥ 0 such that for all n ≥ n0,
c1n3 ≤ 4n3 ≤ c2n3.
Try c1 = c2 = 4; n0 = 1:
4n3 ≤ 4n3 ≤ 4n3.
That was too easy! It's harder when you need to absorb a lower order term:
Show that ∃ c1, c2, n0≥ 0 such that ∀ n ≥ n0,
c1n3 ≤ 4n3 + 2n ≤ c2n3
Try c1 = 4; c2 = 5. (I chose c1 = 4 to make the first term more like the second: certainly adding 2n will make it larger. I chose 5 since the right hand term has to grow larger.)
4n3 ≤ 4n3 + 2n ≤ 5n3
Divide by n3:
4 ≤ 4 + 2/n2 ≤ 5
This is true for all n ≥ n0 = 2, since the term 2/n2 will always be larger than 0 (satisfying the first inequality) but less than 1 (satisfying the second inequality).
You should be able to do similar analyses.