Proving that $f(x)=x^3+x+2$ is bijective without calculus I want to prove that $f(x)=x^3+x+2$, $f: \mathbb R \rightarrow \mathbb R$ is bijective without calculus. My attempts at showing to prove that it' injective and surjective are written below:
$1)$ Injectivity: 
I want to show that $\forall a,b \in \mathbb R$ $f(a)=f(b) \implies a=b$.
I started like this:
$$f(a)=f(b) \implies a^3+a+2=b^3+b+2$$ 
$$\implies a(a^2+1)=b(b^2+1)$$
$$\implies \frac{a}{b}=\frac{b^2+1}{a^2+1}$$
Then I said since $\frac{b^2+1}{a^2+1}>0$ $\forall a,b \in \mathbb R$ then either $a \land b < 0$ or $a \land b > 0$. (For the case when $b=0 \land a \in \mathbb R$ it would be easy to prove that $a=b$.) From there it seemed pretty obvious that $\frac{a}{b}=\frac{b^2+1}{a^2+1} \implies a=b$ so I couldn't really draw a logical argument to show that $a=b$.
$2)$ Surjectivity: 
I want to show that $\forall b \in \mathbb R$ $\exists a \in \mathbb R$ s.t. $f(a)=b$. 
I started like this:
Let $b \in \mathbb R$ and set $f(a)=b$ then we have:
$$a^3+a+2=b$$
$$\implies a^3+a=b-2$$
$$\implies a(a^2+1)=b-2$$
But then I couldn't find an expression for $a \in \mathbb R$ in terms of $b$. 
So I'm wondering if anyone can tell me how I can proceed with my surjectivity and injectivity proofs.

As for the injectivity, if $a>b$, then $a^3>b^3$, so $f(a)=a^3+a+2>b^3+b+2=f(b)$.
As for the surjectivity, there is a formula to find a real solution of an equation of degree 3; it is not very nice, but you can use it. You can find it here
