Show that if $a\neq b$ then $a^3+a\neq b^3+b$

Question

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 !

Answer 1

$$a^3+a=b^3+b\iff a^3-b^3=b-a$$

$$\iff (a-b)\left(a^2+ab+b^2\right)=b-a$$

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 $a,b\in\Bbb R$.

Answer 2

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 $0^2+0*0+0^2=0$.

Answer 3

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.

Answer 4

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?