Find $\frac{1}{-6} \cdot 6 \cdot 7 + 8 \cdot 2 \div 8 \cdot (7-1)$.
This problem involves order of operations, so I need to follow the rules of PEMDAS: parentheses, exponents, multiplication/division, and addition/subtraction. I will start by simplifying the expression inside the parentheses: $7-1 = 6$. Next, I will look for any exponents, but there are none in this problem. Then, I will perform the multiplication and division operations from left to right. I can rewrite the expression as $\frac{1}{-6} \cdot 6 \cdot 7 + 8 \cdot \frac{2}{8} \cdot 6$. The first multiplication, $\frac{1}{-6} \cdot 6$, gives me $-1$. The second multiplication, $-1 \cdot 7$, gives me $-7$. The third multiplication, $8 \cdot \frac{2}{8}$, gives me $2$. The fourth multiplication, $2 \cdot 6$, gives me $12$. Now I have $-7 + 12$, which is a simple addition problem. The final answer is $-7 + 12 = 5$.

