$f(n):=\sum\limits_{d\mid n}\mu(d)\cdot d^2,$ where $\mu(n)$ is the Möbius function. Compute $f(192).$
First, I found all of the divisors of 192 by trial division by primes in ascending order: $D=\{192,96,48,24,12,6,3,2,1,4,8,16,32,64\}$. Then I removed all non-square-free divisors as it will not be contributing to our sum due to the Mobius function. To do this, it suffices to remove all numbers that are divisible by 4. So our remaining set is $\{6,3,2,1\}$.
Thus, written out explicitly, $f(192)=\mu(6)\cdot 6^2 + \mu(3)\cdot 3^2 + \mu(2)\cdot2^2=36+9+4+1=50$.