Solve the following system of equations: $(x-1)^3+2016(x-1)=-1$ and $(y-1)^3+2016(y-1)=1$

Question

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.

Answer 1

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 $a+b=0\implies x+y=2$

Answer 2

Hint

What can be concluded from $$(a+b)(a^2-ab+b^2+2016)=0\\a^2-ab+b^2\ge 0\quad,\quad \forall a,b\in\Bbb R$$?

Answer 3

Consider the graph $a^3+2016a$. This graph has a $pi$ rotational symmetry about the origin. Hence, a real root lets say $r$ of $a^3+2016a+1 $, there also exists $-r$ a real root of $a^3+2016a-1$. Hence, if $x-1$ is equal to $r$ and $y-1$ equal to $-r$ then $x-1+y-1=0$

Edit: It's also clear algebraically that if $r$ solves $a^3+2016a+1=0 $ then $-r$ solves $a^3+2016a-1=0 $