Prove $a^4+b^4+(a-b)^4=c^4+d^4+(c-d)^4$ if $a^2+b^2+(a-b)^2=c^2+d^2+(c-d)^2$ 
if $a,b,c,d$ are positive real numbers and $$a^2+b^2+(a-b)^2=c^2+d^2+(c-d)^2$$
  Prove $a^4+b^4+(a-b)^4=c^4+d^4+(c-d)^4$

Things i have done: from assumption $a^2+b^2+(a-b)^2=c^2+d^2+(c-d)^2$ I can conclude that $$a^2+b^2-ab=c^2+d^2-cd$$
Powering both sides to two gives $$a^4+b^4+a^2b^2-2a^3b-2ab^3=c^4+d^4+c^2d^2-2c^3d-2cd^3$$
And $$a^4+b^4+(a-b)^4=c^4+d^4+(c-d)^4 \rightarrow 2a^4+2b^4+6a^2b^2-2ab^3-2a^3b=2c^4+2d^4+6c^2d^2-2cd^3-2c^3d$$
I can't continue any more.

Hint: 
$$\left(a^2+b^2+(a-b)^2\right)^2=2\left(a^4+b^4+(a-b)^4\right).$$
I wish I knew a smart, non brute-force, way to show that.
