If $\|\mathbf{a}\| = 3$ and $\|\mathbf{b}\| = 6,$ then find $(\mathbf{a} + \mathbf{b}) \cdot (\mathbf{a} - \mathbf{b}).$
I notice that this problem involves dot products, which are related to the cosine of the angle between two vectors. I also recall that the dot product of a vector with itself is equal to the square of its magnitude. This suggests that I might want to use the identity $(\mathbf{x} + \mathbf{y}) \cdot (\mathbf{x} - \mathbf{y}) = \mathbf{x} \cdot \mathbf{x} - \mathbf{y} \cdot \mathbf{y},$ which is a vector version of the difference of squares formula. If I apply this identity to the given expression, I get $(\mathbf{a} + \mathbf{b}) \cdot (\mathbf{a} - \mathbf{b}) = \mathbf{a} \cdot \mathbf{a} - \mathbf{b} \cdot \mathbf{b}.$ Now I can use the fact that $\mathbf{a} \cdot \mathbf{a} = \|\mathbf{a}\|^2 = 9$ and $\mathbf{b} \cdot \mathbf{b} = \|\mathbf{b}\|^2 = 36.$ Therefore, the final answer is $9 - 36 = -27.$ 