<s>[INST] Solve the following system of equations: $(x-1)^3+2016(x-1)=-1$ and $(y-1)^3+2016(y-1)=1$ This problem came up in my homework for an online class and try as I might, I can't find a solution. 
$(x-1)^3+2016(x-1)=-1$
$(y-1)^3+2016(y-1)=1$
Find $x+y$
So far, I've tried letting $(x-1)=a$ and $(y-1)=b$ and then adding the two equations and factoring out $(a+b)$ to get:
$(a+b)(a^2-ab+b^2+2016)=0$
But I don't know where to go from here. 
 [/INST] From your point you have that or $a+b=0$ or $a^2-ab+b^2+2016=0$. But $a^2-ab+b^2+2016=(a-\frac b2)^2 +b^2-\frac{b^2}4+2016>0\forall a,b\in\Bbb R$. So you have $a+b=0\implies x+y=2$
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