Proving Something about Orthogonal Vectors If $\vec x ,\vec y  \in \mathbb R^3$ are orthogonal and $x = \|\vec x\|$ then prove that
$$
\vec x \times \bigl(
  \vec x \times \bigl(
    \vec x \times (
      \vec x \times \vec y
    )
   \bigr)
\bigr)
=
x^4\vec y
$$
I have no idea how to start this, any tips? (Sorry about the $x$'s and multiply signs making it hard to read, that's how it is on my sheet).
 A: Here is a step by step approach
$1.$ You can use the famous indentity
$$\vec{x} \times ( \vec{y} \times \vec{z})=(\vec{x} \cdot \vec{z})\vec{y}-(\vec{x} \cdot \vec{y})\vec{z}$$
$2.$ Next consider that
$$\begin{align}
& \quad \,\,\,\vec{x} \times (\vec{x} \times \vec{y}) \\
&=(\vec{x} \cdot \vec{y})x-(\vec{x} \cdot \vec{x})\vec{y}  \\
&= 0 \vec{x} -x^2 \vec{y} \\
&= -x^2 \vec{y}
\end{align}$$
$3.$ Finally
$$\begin{align}
& \quad \,\,\,\vec{x} \times (\vec{x} \times -x^2\vec{y}) \\
&=-x^2\vec{x} \times (\vec{x} \times \vec{y}) \\
&=-x^2(-x^2 \vec{y}) \\
&= x^4 \vec{y}
\end{align}$$
A: Assume that $\vec{x}$ and $\vec{y}$ are orthogonal. If $\vec{x} = 0$ or $\vec{y} = 0$, the equality is clear. Otherwise, $\vec{x}$ and $\vec{y}$ are linearly independent and $\vec{z} = \vec{x} \times \vec{y}$ is a vector that is perpendicular to both $\vec{x}$ and $\vec{y}$ of length $xy$. Then $\vec{x} \times \vec{z}$ is a vector perpendicular to $\vec{x}$ and $\vec{z}$ of length $x^2y$. This are only two possible vectors that are perpendicular to $\vec{x}$ and $\vec{z}$ of length $x^2y$ - they are $\pm x^2 \vec{y}$. 
Then
$$ \vec{x} \times (\vec{x} \times (\vec{x} \times (\vec{x} \times \vec{y}))) = \vec{x} \times (\vec{x} \times (\pm x^2 \vec{y})) = (\pm x^2) (\vec{x} \times \vec{z}) = x^4 \vec{y}.$$
