Prove $f(x)=x^3+x$ is injective How can algebraically prove $f(x)=x^{3}+x$ is injective.
I can get to $a^2+b^2+ab=-1$ but I can't go any further.
 A: It suffices to note that $f(x)$ is a strictly increasing function on the reals.

That said, if you do want to go that route, we have
$$\begin{align}
f(a) = f(b)
&\iff
a^3+a = b^3 +b
\\&\iff
(a^3-b^3)+(a-b) = 0
\\&\iff
(a-b)(a^2+ab+b^2)+(a-b) = 0
\\&\iff
(a-b)(a^2+ab+b^2+1) = 0
\\&\iff
a = b \,\,\,\text{ or }\,\,\, a^2+ab+b^2+1 = 0
\end{align}$$
Thinking of $a^2+ab+b^2+1 = 0$ as a quadratic equation in $a$, we see that the discriminant is $\Delta = b^2 - 4(b^2+1) = -3b^2 -4 < 0$, so the equation has no solution on the reals.
It follows that the only option is $a=b$, and hence $f$ is injective.
A: $$a^{2} +b^{2}+ ab = \left(a + \frac{b}{2}\right)^{2} + \frac{3b^{2}}{4} \geq 0$$
A: The following solution might seem "overkill", but its interest is that it can be used in other contexts. Let :
$$f(x)=y=x^3+x \tag{1}$$
We have the following diagram :
$$\begin{array}{ccc}
&\mathbb{R} & \xrightarrow{f} & \mathbb{R}& \\
g&\downarrow &  & \uparrow &h \\
&\mathbb{R} & \xrightarrow{F} & \mathbb{R}&
\end{array} \tag{2}$$
where
$$\begin{cases}f&& \ \text{is defined by (1)}\\g(x)&=&\text{arcsinh}(\sqrt{\dfrac{3}{4}}x)=u\\F(u)&=&3u=v\\h(v)&=&w=\dfrac{1}{3}\sqrt{\dfrac{4}{3}} \sinh(v)\end{cases}$$
As a consequence of diagram (2),

$$f=h\circ F \circ g, \tag{3}$$
being a composition of 3 bijections, is itself a bijection.

Explanation :
This is due to the following relationship in hyperbolic trigonometry :
$$\sinh(3a)=4\sinh(a)^3+3\sinh(a)\tag{4}$$
Indeed, setting :
$$\begin{cases}\sinh(a)&=&\sqrt{\dfrac{3}{4}}x\\\sinh(3a)&=&3\sqrt{\dfrac{3}{4}}y\end{cases}\tag{5}$$
and plugging these relationships into (4), we obtain
$$3\sqrt{\dfrac{3}{4}}y=4\dfrac{3}{4}\sqrt{\dfrac{3}{4}}x^3+3\sqrt{\dfrac{3}{4}}x$$
which is equivalent to relationship (1).
At last, from (5) we can extract :
$$ 3a=\text{arcsinh}(3\sqrt{\dfrac{3}{4}}y)=3 \ \text{arcsinh}(\sqrt{\dfrac{3}{4}}x)$$
from which
$$y=\underbrace{\dfrac{1}{3}\sqrt{\dfrac{4}{3}}\sinh(\underbrace{3 \ \underbrace{\text{arcsinh} \sqrt{\dfrac{3}{4}}x)}_u}_v}_w$$
which expresses nothing but relationship (3).
Remark : this method can be plced in correspondence with trigonometric solution of the third degree equation. See for example this.
A: We have $f'(x)=3x^2+1 \ge 1 >0$. Hence $f$ is strictly increasing and therefore injective.
A: The quadratic form $a^2+ab+b^2$ is definite positive, so it can't take the value $-1$.
A: As mentioned, it is enough to show $x^3+x=y^3+y$ implies $x=y$. That leads to  $$(x-y)(x^2+y^2+xy+1)=0.$$
Define the function $f(x,y)=x^2+y^2+xy+1$. 
By the second derivative test, $f$ has a local minimum at $(0,0)$. One can also show that it is also global. That is; $f(x,y)\geq f(0,0)=1$ for all $(x,y)$. Hence $x^2+y^2+xy+1\neq 0$ and $x=y$ as desired.
