Why does $(a-2b)\times (3a+2b) = a\times (3a+2b) - 2b \times(3a+2b)$?

Question

Let $\textbf{a},\textbf{b}\in\mathbb{R}^3$ be such that $|\textbf{a}| = |\textbf{b}|$ and the angle between them is $45º$. We had a test where we should find the answer of $(\textbf{a}-2\textbf{b})\times(3\textbf{a}+2\textbf{b})$. The answer was $8 (\textbf{a}\times \textbf{b})$, but I couldn't understand how: $$ (\textbf{a}-2\textbf{b})\times(3\textbf{a}+2\textbf{b}) = \textbf{a}\times(3\textbf{a}+2\textbf{b}) - 2\textbf{b}\times(3\textbf{a}+2\textbf{b}) \\= 3 (\textbf{a}\times \textbf{a}) + 2 (\textbf{a}\times \textbf{b}) - 6 (\textbf{b}\times \textbf{a}) - 4 (\textbf{b}\times \textbf{b}) = 8 (\textbf{a}\times \textbf{b}) $$ Can anybody explain the part above?

Answer 1

Use the fact that $a\times a=0$, $b\times b=0$ and $a\times b=-b\times a$. I don't think you need $|a|=|b|$ for this proof.