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If $n=q_1 \dots q_k$, where $q_j$ are prime powers of distinct prime numbers, then computing $\mu(n)=\mu(q_1 \dots q_k)$ is equivalent to computing $\mu(q_1)\dots \mu(q_k)$. To know that $\mu(n)=\mu(q_1)\dots \mu(q_k)$, you have to know that $n=q_1 \dots q_k$ and that the $q_j$'s are prime powers of distinct prime numbers. This is the same as computing the prime factorization of $n$.