Vector dot products

Question

Consider two vectors $\vec{a\:}$, $\vec{b\:}$ with $|\vec{a\:}| = 2, > |\vec{b\:}| = 1$, and an angle between them of $60^\circ$ . Find the scalar product$$\left(\vec{a\:}+\vec{b\:}\right)\cdot \left(\vec{a\:}-\vec{b\:}\right)$$

So I know that $\left(\vec{a\:}+\vec{b\:}\right)\cdot \left(\vec{a\:}-\vec{b\:}\right)= \left|\vec{a\:}\right|^2-\left|\vec{b\:}\right|^2$

But in the solution it says that it is equal to $$\left|\vec{a\:}\right|^2-\left|\vec{a\:}\right|\left|\vec{b\:}\right|cos\:60-2\left|\vec{b\:}\right|^2$$

Why though?

edit: Here are the pics

Answer 1

the expression is given $$(\vec a + \vec b).(\vec a - \vec b)=\vec a.\vec a-\vec a.\vec b+\vec b.\vec a-\vec b.\vec b$$ $$=|\vec a|.|\vec a|\cos 0^o-|\vec a|.|\vec b|\cos 60^o+|\vec a|.|\vec b|\cos 60^o-|\vec b|.|\vec b|\cos 0^o$$ $$=|\vec a|.|\vec a|-|\vec b|.|\vec b|$$ $$=|\vec a|^2-|\vec b|^2$$

Answer 2

Note that you ask a question about $$ (a+b) (a-b) $$ but the question in the book is about $$ (a+b) (a-2b) $$ I think you missed the factor of $2$.