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.

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.
