Do it logically

We know that $2n$ will be even .Therefore $2n+1$ will be odd .Let $a^{\color{#20A900}{2n+1}}$ be $a^{\color{#20A900}{k}}$ and similarly $b^{\color{#20A900}{2n+1}}=b^{\color{#20A900}{k}}$ .Now $a^{\color{#20A900}{k}}+b^{\color{#20A900}{k}}$ can be factorised as $(a+b)(a^{\color{#20A900}{k}-1}-a^{\color{#20A900}{k}-2}b+\ldots+ab^{\color{#20A900}{k}-2}-b^{\color{#20A900}{k}-1}).\quad\quad\quad\quad\quad\quad\quad\quad\quad[k \text{ is odd.}]$ Therefore we have shown that $a+b$ is a factor of $a^{\color{#20A900}{k}}+b^{\color{#20A900}{k}} ~\text {or} ~a+b|a^{2n+1}+b^{2n+1}$ .

If we substitute n=1 in the eqn we get a^3 + b^3.so why not option b