2015/085) Prove that

$(x+y)^3 + (y+z)^3 + (z+x)^3 -3(x+y)(y+z)(z+x)=2(x^3+y^3+z^3 – 3xyz)$
Solution 
knowing
$a^3+b^3 + c^3 – 3abc = \frac{1}{2}(a+b+c)((a-b)^2 + (b-c)^2 + (c-a)^2)\cdots (1)$
putting $a = x + y , b= y + z, c= z+x$ we get

$(x+y)^3 + (y+z)^3 + (z+x)^3 -3(x+y)(y+z)(z+x)$

= $\frac{1}{2}(2x+2y+ 2z)((x-z)^2 + (y-x)^2 + (z-y)^2)$

= $2 * \frac{1}{2}(x+y+z)((x-y)^2+(y-z)^2 + (z-x)^2)$

= $2(x^3+y^3+z^3 – 3xyz)$ using (1)