<s>[INST] Prove $a^3+b^3+c^3+d^3=3(abc+bcd+cda+dab)$ 
If $a,b,c,d$ are real numbers and $a+b+c+d=0$. prove:
$$a^3+b^3+c^3+d^3=3(abc+bcd+cda+dab)$$

Using Euler's identity: if $x+y+z=0$ then $x^3+y^3+z^3=3xyz$.  in this problem substitute $x=a+b,y=c,z=d:$
$$(a+b)^3+c^3+d^3=3cd(a+b)$$
$$a^3+b^3+c^3+d^3=3(a+b)(cd-ab)=3(acd-a^2b+bcd-ab^2)$$
Although I am sure my approach is right, but I should have $3(abc+bcd+cda+dab)$ at the RHS. I don't know how to get it.
 [/INST] $$(a+b)^3=a^3+b^3+3ab(a+b)=a^3+b^3-3ab(c+d)$$
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