Prove $a^4+b^4+(a-b)^4=c^4+d^4+(c-d)^4$ if $a^2+b^2+(a-b)^2=c^2+d^2+(c-d)^2$

Question

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.

Answer 1

Observe that: $$ a^4+b^4+(a-b)^4-c^4-d^4-(c-d)^4\\ =2(a^2-ab+b^2-c^2+cd-d^2)(a^2-ab+b^2+c^2-cd+d^2) $$ so the result follows because you have proved the first term on the RHS is $0$.

Answer 2

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.