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: $$a^3+a=b^3+b\iff a^3-b^3=b-a$$
$$\iff (a-b)\left(a^2+ab+b^2\right)=b-a$$
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 $a,b\in\Bbb R$.
A: Using analysis: define $f(x)=x^3+x$. Since $f'(x)=3x^2+1>0$, the function is strictly increasing, hence, injective, and the result follows.
A: 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 $0^2+0*0+0^2=0$.
A: I'm not a math major so I'm not sure about the formalism but does this work?
If a!=b,
Then a=b+d ; where d is a non zero integer.
Plug in to the expression (a^3+a)-(b^3+b) :-
3bd^2+3db^2+d^3.
This has only one real root which is d=0, which is a contradiction, so the two expressions cannot be equal. 
Too informal? Correct or incorrect, math majors?
