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 !

$$a^3+a=b^3+b\iff a^3-b^3=b-a$$
$$\iff (a-b)\left(a^2+ab+b^2\right)=b-a$$
If $a=b$, then we're done. For contradiction, assume $a\neq b$. Then $a-b\neq 0$ and we can divide both sides by $a-b$:
$$a^2+ab+b^2=-1\iff 4a^2+4ab+4b^2=-4$$
$$\iff (2a+b)^2+3b^2=-4,$$
contradiction, because $(2a+b)^2+3b^2\ge 0$ for all $a,b\in\Bbb R$.
