Prove that $f(x):\mathbb{R}\to\mathbb{R}$ , $x \mapsto x^3$ is injective. Prove that $f(x):\mathbb{R}\to\mathbb{R}$ ,  $x \mapsto x^3$ is injective.

I want to prove this claim is true. 
Here is my outline so far:

We want to show that $f(a)=f(b)$ implies that $a=b$, for all $a,b \in \mathbb{R}$
We have $f(a)=a^3$, and $f(b)=b^3$
So, if $f(a)=f(b)$, we have $a^3=b^3$, or $a^3-b^3=0$, or $(a-b)(a^2+ab+b^2)=0$
Consider two cases:
Case 1: $(a-b)=0$, then $a=b$, and we are done.
Case 2: $(a^2+ab+b^2)=0$ We want to show that $a=b$.

I am having trouble with Case 2. I need to figure out how to show that $a=b$. I think I can use this inequality some how: $a^2+2ab+b^2>0$ 

Any help would be appreciated.
 A: Think of it as a quadratic on $a$: if we complete the square,
$$
a^2+ab+b^2=\left(a+\frac b2\right)^2+b^2-\frac {b^2}4=\left(a+\frac b2\right)^2+\frac{3b^2}4.
$$
For this to be zero we need both summands to be zero (because both are nonnegative). Then we get first that $b=0$, and then that $a^2=0$, so $a=0$. 
So, for $a,b$ not both zero, 
$$
a^2+ab+b^2>0.
$$
A: Here is an alternative suggestion for a proof. First, show that $f$ is strictly increasing. Consider $x\in\mathbb{R}$, and define some $\epsilon>0$.
$$ f(x+\epsilon)=(x+\epsilon)^3=x^3+3x^2\epsilon+3x\epsilon^2+\epsilon^3>x^3=f(x),\forall x\in\mathbb{R}. $$
Now, suppose this function were not injective, that is, there existed some $y,y'\in \mathbb{R}$ such that $f(y)=f(y')$ but $y\ne y'$. Because $y\ne y'$, by trichotomy of the reals, $y>y'$ or $y'>y$. Assume w.l.o.g. that $y'>y$. Then we have $y'>y$, but $f(y')\not>f(y)$. This contradicts $f$ strictly increasing. $f$ is injective. $\square$
A: Your strategy works fine. You do not need to show that the function is increasing (although, that would be one way to do it).
You just need to show that $a^2+ab+b^2=0$ does not hold for any $a\neq b$. To this end, fix $b$ and try to use the quadratic formula to find whether there is a value of $a$ that satisfies the equation. Spoiler alert: $\Delta<0$.
