Prove $n^3 + 7n + 3$ is divisible by 3 for all integers n ≥ 0 The statement I'm trying to prove is:

$n^3 + 7n + 3$ is divisible by 3 for all integers n ≥ 0  

I eventually need to prove $(k + 1)^3 + 7(k + 1) + 3$ is divisible by 3.

I don't really understand how to deal with $k + 1$, so I'm a little lost.

I've know that the base case of P(0) is true, but I'm not sure about proving the inductive case.
The statement is not true. Take $n=1$ as a counter example.     

Since it's not true, you won't manage to prove it (by induction or otherwise).
The given statement  $n^3 + 7n + 3$  is divisible by 3 for  $n ≥ 0$ is not true, as detailed below
P(n): n^3. + 7n + 3  is divisible by 3
P(1): 1^3 + 7(1) + 3 = 11 which is not divisible by 3
P(2): 2^3 + 7(2) + 3 = 25 which is also not divisible by 3
Hence given statement is not true.

I presume that statement may be n^3 - 7n + 3  is divisible by 3 for n ≥ 0
In that case Proof is detailed below.

P(n):   n^3 - 7n + 3   is divisible by 3
        P(1):   (1)^3 -7(1) + 3 = 1 -7 + 3 = -3 which is divisible by 3

Assume P(k) is true for some positive integer k, i.e 
P(k):  (k)^3 - 7k + 3   is divisible by 3
       We can write  (k)^3 - 7k + 3   = 3d, where d ∈ N ………….. (1)
       We shall prove that P(k+1) is also true, whenever P(k) is true. Now we have
       P(k+1):   (k+1)^3 - 7(k+1) + 3
        = k^3 + 1^3 +3.k.1(k+1) - 7(k+1) + 3 (Using (a+b)^3 = a^3 + b^3 + 3ab(a+b) 

    = k^3 + 1 +3k(k+1) - 7(k+1) + 3  
    = k^3 + 1 + 3k^2 + 3k -7k - 7 +3
    = k^3 - 7k + 3 + 3k^2 +3k - 6  (by rearranging)
    = 3d + 3(k^2 + k - 2 ).........{From (1)}      
    = 3d + 3(k^2 + 2k - k - 2 ) 
    = 3d + 3 {k(k +2)-(k+2)}
    = 3d + 3 (k-1)(k+2)
    = 3 {d +(k-1)(k+2)}
    = 3 q,  where q = {d +(k-1)(k+2)} which is some positive integer


From the last line, we see  (k+1)^3 - 7(k+1) + 3 ) is divisible by 3.
Thus P(k+1) is true, whenever P(k) is true.
Hence, by the PMI,  the given statement is true for all natural numbers n ≥ 0