Show that if $a\neq b$ then $a^3+a\neq b^3+b$ Show that if $a\neq b$ then $a^3+a\neq b^3+b$
We assume that $a^3+a=b^3+b$ to show that $a=b$
$$\begin{align}
a^3+a=b^3+b &\iff a^3-b^3=b-a\\
&\iff(a-b)(a^2+ab+b^2)=b-a\\
&\iff a^2+ab+b^2=-1
\end{align}$$
Im stuck here !

If you have calculus, you can conclude this is impossible as follows, since $lim_{x, y \to \infty}x^2+xy+y^2=\infty$, there is a point where the minima is achieved. This point has vanishing partial derivatives, thus $\partial_x(x^2+xy+y^2)=2x+y=0$ and likewise $2y+x=0$, so $x=-2y=4x$ so $x=0$ and likewise $y=0$, so $x=y=0$, and we have a contridiction since $0^2+0*0+0^2=0$.
