Remember that steps to induction goes like this
Let P be a statement. Show that
P(1) is true
If P(k) is true, then P(k+1) is true
You already showed that P(1) is true. Now just need to show P(k+1) is true if P(k) is true. Let P(k) be true. Then k^3 + 5k is divisible by 6. Need to show that (k+1)^3 + 5(k+1) is divisible by 6. Expand.
(k+1)^3 + 5(k+1) = k^3 + 3k^2 + 3k + 1 + 5k + 5 = (k^3 + 5k) + 3k(k+1) + 6.
Clearly 6 is divisible by 6. By induction assumption k^3 + 5k is divisible by 6. I claim that 3k(k+1) is divisible by 6. That is because k(k+1) is even for all k. If k is even, then clearly k(k+1) is even so 3k(k+1) is divisible by 6. If k is odd, then k + 1 is even so k(k+1) is even and again 3k(k+1) is divisible by 6. The sum of terms each of which are divisible by 6 is also divisible by 6.
I know the OP probably needed to use induction. For completeness, I'll give a proof without induction. I hope this gives more intuition as to why the statement is true.
Lemma
is divisible by 6 if and only if is divisible by 6.
Proof: Suppose , then . because since for . Conversely, .
Proposition
is divisible by 6.
Proof: From the lemma, the proposition holds iff is divisible by 6. But . These are three consecutive integers, and so one of them must be divisible by 3.