a+b+c+d=0
a+b=-(c+d)
a^3+b^3+c^3+d^3
=(a+b)(a^2-ab+b^2)+(c+d)(c^2-cd+d^2)
=(a+b)[(a+b)^2-3ab]+(c+d)[(c+d)^2-3cd]
=(a+b)^3-3ab(a+b)+(c+d)^3-3cd(c+d)
=3ab(c+d)-3cd(c+d)=3
(ab-cd)(c+d)=1
abc+bcd+acd+abd
=ab(c+d)+cd(a+b)
=ab(c+d)-cd(c+d)
=(ab-cd)(c+d)
=1