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

Let $f:\mathbb{R}\to\mathbb{R}$ defined by $f(x)=x^3+3x+1$, let's suppose $x$ and $y$ are real numbers, then
\begin{align*}
f(x)=f(y)\quad \iff\quad x^3+3x+1&=y^3+3y+1\\
x^3-y^3+3(x-y)&=0
\end{align*}
We can factor $x^3-y^3$ as $(x-y)(x^2+xy+y^2)$, so
\begin{align*}
f(x)=f(y)\quad \iff(x-y)(x^2+xy+y^2+3)&=0
\end{align*}
Since $$x^2+xy+y^2=(x+\frac{1}{2}y)^2+\frac{3}{4}y^2\ge 0$$ we have $$x^2+xy+y^2+3\ge 3$$
Then
\begin{align*}
f(x)=f(y)\quad \iff x-y&=0 \iff x=y
\end{align*}
