This problem involves a custom operation, so I need to follow the definition given in the problem. To find $6 \& 3$, I need to plug in $a=6$ and $b=3$ into the formula $a \& b = (a+b)(a-b)$. Simplifying, I get $6 \& 3 = (6+3)(6-3) = 9 \times 3 = 27$. 
The operation $\&$ is defined as $a \& b = (a+b)(a-b)$. What is the value of $6 \& 3$?