$(x+y+z)^3-(y+z-x)^3-(z+x-y)^3-(x+y-z)^3=24xyz$? The question given is 

Show that $(x+y+z)^3-(y+z-x)^3-(z+x-y)^3-(x+y-z)^3=24xyz$.

What I tried is suppose $a=(y+z-x),\ b=(z+x-y)$ and $c=(x+y-z)$ and then noted that $a+b+c=x+y+z$. So the question statement reduced to $(a+b+c)^3-(a^3+b^3+c^3)$. Then I tried to invoke the identity $(a^3+b^3+c^3-3abc)=(a+b+c)(a^2+b^2+c^2-ab-bc-ac)$ by adding and subtracting $3abc$ in the question statement. After doing all this when I substituted back the values of $a,b$ and $c$, I ended up with the initial question statement.
Any hints will be appreciated.
 A: HINT:
Following your way, 
$$(a+b+c)^3-(a^3+b^3+c^3)=3(a+b)(b+c)(c+a)$$
A: Put $z = 0$, the LHS vanishes.  So $z$ must be a factor.  By symmetry $xyz$ must be a factor.  As LHS is of third degree, the only other factor must be a constant, try $x=y=z=1$ to get that...
A: Observe that,
$\begin{align}(x+y+z)^3-(x+y-z)^3&=2z\left((x+y+z)^2+(x+y-z)^2)+(x+y)^2-z^2\right)\\&=2z\left(3(x+y)^2+z^2\right)\tag{1}\end{align}$
and,
$\begin{align}(x-y-z)^3-(x-y+z)^3&=-2z\left((x-y-z)^2+(x-y+z)^2)+(x-y)^2-z^2\right)\\&=2z\left(3(x-y)^2+z^2\right)\tag{2}\end{align}$
Therefore,
$(x+y+z)^3+(x-y-z)^3-(x-y+z)^3-(x+y-z)^3\\=\bigl((x+y+z)^3-(x+y-z)^3\bigr)+\bigl((x-y-z)^3-(x-y+z)^3\bigr)\\=2z\left(3(x+y)^2+z^2\right)-2z\left(3(x-y)^2+z^2\right)\\=2z(3\cdot 4xy)\\=24xy$
A: Pascal's (or Tartaglia's) Tetrahedron: the left outline is a binomial expansion of $(x+y)^3$, while the right outline is a binomial expansion of $(x+z)^3$ and the bottom outline is a binomial expansion of $(y+z)^3$. Each of them is highlighted in yellow for identification purposes. Furthermore, you may notice that the terms with the highest powers are situated at the vertices, because the coefficients of those terms are “1”. Imagine to rewrite this equation in "graphically", treating expressions in triangles as the components of a matrix (which add-term contracts):

Hence the claim.
