The question states:
Suppose $ a,b \in \Bbb R^3$. Show that
$$ (a \times a) \times b = a \times (a \times b) $$
if and only if $a$ and $b$ are proportional (i.e. one is a scalar multiple of the other).
If we assume that one is a scalar multiple of the other, than proving the equation is easy, using various properties such as $a \times a = 0$ and $a \times ka = k (a \times a).$ However, I couldn't really figure out how to deduce $a$ and $b$ are proportional if the above equation is true. I get that the left side of the equation is zero. From this, I deduced that
$$ a \times b = ka. $$ However, I couldn't progress much further than this. Can anyone help?
As you've mentioned, we have $a\times (a\times b) = 0$. This occurs if and only if $a$ is parallel to $a\times b$ or if $a\times b = 0$. If $a\times b = 0$ then they're scalar multiples and we're done. The second case cannot happen: if $a$ and $b$ are not parallel then $a \times b$ is orthogonal to $a$, so $a\times (a\times b)$ is not $0$.