Simplify this vector expression I have a question which I know the answer to but am having difficulty showing it. It's about simplifying the following vector equation (I'm aware I've grouped the terms in an arguably strange way):  
$ (p^2 + q^2 |\mathbf{P}|^2 + (\mathbf{P} \bullet \mathbf{Q})^2)\mathbf{X} + (\mathbf{Q} \bullet \mathbf{X})(|\mathbf{P}|^2 \mathbf{Q} - (\mathbf{P} \bullet \mathbf{Q})\mathbf{P}) + (\mathbf{P} \bullet \mathbf{X})(|\mathbf{Q}|^2 \mathbf{P} - (\mathbf{P} \bullet \mathbf{Q})\mathbf{Q}) + (\mathbf{X} \bullet (\mathbf{P} \times \mathbf{Q}))(\mathbf{P} \times \mathbf{Q}), $
where $p$ and $q$ are scalars and $\mathbf{P} , \mathbf{Q} , \mathbf{X} $ are three dimensional vectors equipped with the standard dot, $\bullet$, and cross, $\times$, products. Also, $|\mathbf{P}|^2 \equiv (\mathbf{P} \bullet \mathbf{P}) $.
Furthermore, we have the condition that $p^2 + |\mathbf{P}|^2 = q^2 + |\mathbf{Q}|^2 = 1 $.
I'm quite sure that this expression should just simplify to $\mathbf{X}$, particularly having run different values of each variable and vector through some code. Maybe the last three terms always linearly combine to be parallel to $\mathbf{X}$?
Any help would be much appreciated! 

Many thanks for the reply. After a few hours straight of trying to tackle this, I finally solved the problem. I'm going to use these two vector identities (which I've pulled from Wikipedia and trust to be correct).
$ (\mathbf{A} \times \mathbf{B}) \bullet (\mathbf{C} \times \mathbf{D}) = (\mathbf{A} \bullet \mathbf{C})(\mathbf{B} \bullet \mathbf{D}) - (\mathbf{B} \bullet \mathbf{C})(\mathbf{A} \bullet \mathbf{D}) \qquad (1)  $
$ \mathbf{D} = \frac{\mathbf{D} \bullet (\mathbf{B} \times \mathbf{C})}{[\mathbf{A},\mathbf{B},\mathbf{C}]}\mathbf{A} + \frac{\mathbf{D} \bullet (\mathbf{C} \times \mathbf{A})}{[\mathbf{A},\mathbf{B},\mathbf{C}]}\mathbf{B} + \frac{\mathbf{D} \bullet (\mathbf{A} \times \mathbf{B})}{[\mathbf{A},\mathbf{B},\mathbf{C}]}\mathbf{C} \qquad (2)  $
Where $[\mathbf{A},\mathbf{B},\mathbf{C}] = \mathbf{A} \bullet (\mathbf{B} \times \mathbf{C}) $ and $(2)$ describes how to describe a vector $\mathbf{D}$ in terms of three arbitrary vectors $\mathbf{A},\mathbf{B},\mathbf{C}$.
So let's begin!
So first, I rearrange the expression into a more useful form:
$ (p^2 + q^2  |\mathbf{P}|^2 + (\mathbf{P}\bullet\mathbf{Q})^2)\mathbf{X} + ((\mathbf{Q}\bullet\mathbf{X})|\mathbf{P}|^2 - (\mathbf{P}\bullet\mathbf{X})(\mathbf{P}\bullet\mathbf{Q}))\mathbf{Q} + ((\mathbf{P}\bullet\mathbf{X})|\mathbf{Q}|^2 - (\mathbf{Q}\bullet\mathbf{X})(\mathbf{P}\bullet\mathbf{Q}))\mathbf{P} +(\mathbf{X}\bullet(\mathbf{P}\times\mathbf{Q}))\mathbf{P}\times\mathbf{Q}  $
Using relation $(1)$, we can rewrite the coefficients of $\mathbf{P}$ and $\mathbf{Q}$ such that we have
$ (p^2 + q^2  |\mathbf{P}|^2 + (\mathbf{P}\bullet\mathbf{Q})^2)\mathbf{X} - ((\mathbf{P}\times\mathbf{Q})\bullet(\mathbf{X}\times\mathbf{P}))\mathbf{Q} + ((\mathbf{P}\times\mathbf{Q})\bullet(\mathbf{X}\times\mathbf{Q}))\mathbf{P} +(\mathbf{X}\bullet(\mathbf{P}\times\mathbf{Q}))\mathbf{P}\times\mathbf{Q}  $
Now, we use relation $(2)$ to rewrite $\mathbf{P}\times\mathbf{Q}$ in terms of $\mathbf{X} , \mathbf{P}$ and $\mathbf{Q}$ such that 
$ \mathbf{P}\times\mathbf{Q} = \frac{(\mathbf{P}\times\mathbf{Q}) \bullet (\mathbf{P} \times \mathbf{Q})}{\mathbf{X}\bullet(\mathbf{P}\times\mathbf{Q})}\mathbf{X} + \frac{(\mathbf{P}\times\mathbf{Q}) \bullet (\mathbf{Q} \times \mathbf{X})}{\mathbf{X}\bullet(\mathbf{P}\times\mathbf{Q})}\mathbf{P} + \frac{(\mathbf{P}\times\mathbf{Q}) \bullet (\mathbf{X} \times \mathbf{P})}{\mathbf{X}\bullet(\mathbf{P}\times\mathbf{Q})}\mathbf{Q} = \frac{(\mathbf{P}\times\mathbf{Q}) \bullet (\mathbf{P} \times \mathbf{Q})}{\mathbf{X}\bullet(\mathbf{P}\times\mathbf{Q})}\mathbf{X} - \frac{(\mathbf{P}\times\mathbf{Q}) \bullet (\mathbf{X} \times \mathbf{Q})}{\mathbf{X}\bullet(\mathbf{P}\times\mathbf{Q})}\mathbf{P} + \frac{(\mathbf{P}\times\mathbf{Q}) \bullet (\mathbf{X} \times \mathbf{P})}{\mathbf{X}\bullet(\mathbf{P}\times\mathbf{Q})}\mathbf{Q}$
Substituting this back, we find that terms cancel and we're left with 
$ (p^2 + q^2  |\mathbf{P}|^2 + (\mathbf{P}\bullet\mathbf{Q})^2 + |\mathbf{P}\times\mathbf{Q}|^2)\mathbf{X} = (p^2 + q^2  |\mathbf{P}|^2 +|\mathbf{P}|^2 |\mathbf{Q}|^2)\mathbf{X} = \mathbf{X} $
Many thanks for the help, really appreciated it; hope you guys don't feel like I've wasted your time... do let me know if you agree with what I've written!
