Let $a$ be the number of positive multiples of $6$ that are less than $30$. Let $b$ be the number of positive integers that are less than $30$, and a multiple of $3$ and a multiple of $2$. Compute $(a - b)^3$.
I notice that both $a$ and $b$ count the number of multiples of $6$ that are less than $30$, since $6$ is the least common multiple of $3$ and $2$. Therefore, $a = b$. This means that $(a - b)^3 = 0^3 = 0$. 