Connection between Möbius and 𝜔 function

How can we from $\sum_{n\leqslant x}\mu (n)=o(x)$ deduce $\sum_{n\leqslant x}(-1)^{\omega(n)}=o(x)$, where $\omega(n)$ is the number of different primes dividing $n$ and$\mu (n)$ is the Möbius function?Heuristic: Up to x there must be roughly the same portion of numbers with odd and even number of prime divisors. First equality ( equivalent to prime number theorem ) prooves that for square free numbers, and there is no reason why should it not be valid for all numbers