Two proofs I'm having difficulty with I've been given an assignment. Almost done except the last two are tripping me up. They are as follows:
1) if $2x^2-x=2y^2-y$ then $x=y$
2) if $x^3+x=y^3+y$ then $x=y$
I imagine they use a similar tactic as they both involve powers, but I've tried factoring,completing the square, difference of squares and difference of cubes and nothing seems to help.
Any hints would be appreciated.

Question $1$: The equation is equivalent to $2x^2-2y^2=x-y$. The left-hand side factors as $2(x+y)(x-y)$. So our equation can be rewritten as
$$2(x+y)(x-y)=x-y.$$ 
Thus  any pair $(x,y)$ such that $x\ne y$ and $2(x+y)=1$ is a counterexample to the assertion that $x$ must be equal to $y$. This was pointed out by ncmathsadist. 
Edit: It turns out that one is supposed to show that the only integer solutions have $x=y$.  This follows from the above calculation, since $2(x+y)=1$ has no integer solutions.
Question $2$: We look at the question as amended. We can factor and rewrite the equation as $(x-y)(x^2+xy+y^2)=-(x-y)$. If $x\ne y$, we can cancel $x-y$, and obtain $x^2+xy+y^2=-1$. 
But the equation $x^2+xy+y^2=-1$ has no real solutions, since $x^2+xy+y^2\ge 0$ for all real $x$ and $y$.  One way to see this is to complete the square, getting 
$$x^2+xy+y^2=\left(x+\frac{y}{2}\right)^2+\frac{3}{4}y^2.$$
The right-hand side is  clearly never negative for real $x$ and $y$. So it is true that the only solutions of the original equation have $x=y$. 
