How to show algebraically that $x^3 +3x +1$ is injective? 
How to show algebraically that $$x^3 +3x +1$$ is injective?

Working with the usual method of assuming that $f(c)=f(d)$ and then seeing if $c=d$. I've tried several approaches, including factoring by the difference of cubes, followed by use of the quadratic formula. I'm stumped. Any help greatly appreciated!  
Thanks, Bob

Notice $$\begin{align}
f(x) - f(y) &= (x^3 + 3x + 1) - (y^3 + 3y+1)\\
&= (x-y)(x^2 + xy + y^2 + 3)\\
&= \frac12 (x-y)( x^2 + (x+y)^2 + y^2 + 6)\end{align}$$
Since $x^2 + (x+y)^2 + y^2 + 6 \ne 0$ for any pair of real numbers $x, y$;
we have $f(x) \ne f(y)$ whenever $x \ne y$.
