<s>[INST] 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.
 [/INST] 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.
</s>