Q.Prove that $n^3 + 2n$ is divisible by $3$ for every natural number $n$.

[!TLDR] Write $(k+1)^3+2(k+1)$ as $(k^3+2k)+3(k^2+k+1)$. [!ANSWER] Proved: $n^3+2n$ is divisi …

Let $P(n)$ be the statement that $n^3+2n$ is divisible by 3. Base case: $n=1$: $1^3+2(1)=1+2=3$, divisible by 3, so $P(1)$ holds. Inductive step: assume $P(k)$: $k^3+2k=3m$ for some integer $m$. Expand: $(k+1)^3+2(k+1)=k^3+3k^2+3k+1+2k+2=(k^3+2k)+3k^2+3k+3=(k^3+2k)+3(k^2+k+1)$. Substituting the inductive hypothesis: $(k+1)^3+2(k+1)=3m+3(k^2+k+1)=3[m+k^2 …

Divisibility-by-induction: expand $(k+1)^3+2(k+1)$, regroup as $(k^3+2k)+3(k^2+k+1)$, an …

Collecting the leftover terms $3k^2+3k+3$ incorrectly and failing to notice all three terms already share the common factor 3, which …