Suppose $ 7\; |\; 3^{n}+n^{3}$. This implies $ 7$ does not divide $ n$. This, in turn, implies that $ n^{6}= 1 (mod \; 7)$. $ 7\; |\; 3^{n}+n^{3}\;\Rightarrow \;7\; |\; n^{3}(3^{n}+n^{3}) \;\Rightarrow \;7\; |\; n^{3}3^{n}+n^{6}\;\Rightarrow\; 7\; |\; n^{3}3^{n}+1+n^{6}-1\; \Rightarrow \;7\; |\; n^{3}3^{n}+1$ (since $ 7\;|\; n^{6}-1$). Suppose $ 7\; |\; n^{3}3^{n}+1$. This implies $ 7$ does not divide $ n$ which, in turn, implies that $ n^{6}= 1 (mod \; 7)$. $ 7\; |\; n^{3}3^{n}+1 \;\Rightarrow \;7\; |\; n^{3}(n^{3}3^{n}+1) \;\Rightarrow \;7\; |\; n^{6}3^{n}+n^{3}\;\Rightarrow\; 7\; |\; (n^{6}-1)3^{n}+3^{n}+n^{3}\; \Rightarrow \;7\; |\; 3^{n}+n^{3}$ (since $ 7\;|\; n^{6}-1$). Hence, $ 7\; |\; 3^{n}+n^{3}$ if and only if $ 7\; |\; n^{3}3^{n}+1$.

Peter wrote: Prove that for every $ n\in\mathbb{N}$ the following proposition holds: $ 7|3^{n}+n^{3}$ if and only if $ 7|3^{n}n^{3}+1$. One may generalize the result slightly. Proposition. Let $ p$ be an odd prime. Suppose that $ m$ and $ n$ are positive integers with $ p\not\vert m$. Then, $ p\;\vert\; n^{\frac{p-1}{2}}+m$ if and only if $ p\;\vert\; n^{\frac{p-1}{2}}m+1$. The proof is almost same with the above solution. One may also check that Propostion. Suppose that $ a^{2}\equiv 1\; (mod\; n)$. Then, $ n\;\vert\; a+b$ if and only if $ n\;\vert\; ab+1$. Outline of Proof. Since $ a^{2}\equiv 1\; (mod\; n)$, it is clear that $ gcd(a, n) = 1$. Now, use the identity $ ab+1 = a(a+b)-(a^{2}-1)$.

Simply use that $n^3\equiv \pm 1 (mod~7)$. This works for the generalisation as well.

easy to see $(n,7)=1$ \[7|3^n+n^3 \iff 7|n^3(3^n+n^3) \iff 7|3^nn^3+1\]