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.

For 1) Observe that (I am assuming $x,y$ are real numbers):
\begin{align}
2x^2-x=2y^2-y\\
&\implies (2x^2-x)-2y^2+x=(2y^2-y)-2y^2+x\\
&\implies 2(x^2-y^2)=(x-y)\\
&\implies 2(x-y)(x+y)=(x-y)\\
&\implies 2(x-y)(x+y)-(x-y)=(x-y)-(x-y)=0\\
&\implies 2(x-y)(x+y)-(x-y)=0\\
&\implies (x-y)(2(x+y)-1)=0\\
&\implies (x-y)=0 \text{ or } 2(x+y)-1=0\\
\end{align} So, either $(x-y)=0$ or $2(x+y)-1=0.$ If $x-y=0$ then $x=y$ and we have our result. What if $2(x+y)-1=0$? Then there is a possiblity that $x\neq y.$ But if $x=y=1/4$ then $2(x+y)-1=2(1/4+1/4)-1=2(1/2)-1=1-1=0.$ Thus in either case $x=y$ is a solution. Like ncmathsadist pointed out $x=1/2$ and $y=0$ also works but that doesn't mean that $x=y$ doesn't work! I think for precalculus level this is enough. 
For 2) do a similar analysis and Brian has done enough for you in his answer.
