Showing that $a \times [a \times [a \times (a \times b)]]=x^4 b$ where $x = \| a \|$

Let $a,b \in \mathbb R^3$ be orthogonal vectors and set $x=\| a \|$. Show that: $a \times [a \times [a \times (a \times b)]]=x^4 b$

So I'm currently here and don't know where to go next:

$a \times (a \times b) = (a \cdot a)b-(a \cdot b)a$

=$(a \cdot a)b - 0$ as $a$,$b$ is orthogonal the dot product = 0

= $(a \cdot a) b$

Left with : $a\times[a\times[(a \cdot a) b]]=x^4b$

Solution 1:

Assuming $x=\|a\|$, you can do the following. First, as you proved: $a\times (a\times b)=-x^2b$ (there is a slight misprint in the question). Then denote $b_1=a\times (a\times b)$, it is also orthogonal to $a$, so applying what you have to $b_1$, you get that your vector is $a\times (a\times b_1)=-x^2b_1=(-x^2)(-x^2)b=x^4b$.

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