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.
 A: 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
A: Injectivity is clear and so is for the fact that there is not a positive root. 
Applying now Descartes's Rule we have 
$$f(x)=x^3+x+2 \text{ have no change signs }\\f(-x)=-x^3-x+2 \text{ has one change sign }$$
Consequently the maximum number of negative roots is $0+1=1$ and $f^{-1}$ exists.
A: 1) Injective:
$f(x) = x^3 +x+2$,  $x \in \mathbb{R}$ is strictly monotonically increasing.
Let $x_1< x_2 $, then
$x_1^3 + (x_1+2)  \lt x_2^3 + (x_2 +2)$,
is strictly monotonically increasing,
sum of $y_1= x^3$ and a linear function $y_2= x+2$, both  strictly monotonically  increasing.
2) Surjective.
A)$\lim_{x \rightarrow \infty} f(x) = \infty.$
B)$\lim_{x \rightarrow -\infty} f(x) = - \infty$.
Let $c \in \mathbb{R}$.
Choose $a$ with $f(a) < c$, and $b$ with $f(b) >c$, 
(property A,B).
Consider the closed interval $ [a,b] .$
Since $f$ is continuos, and $f(a) < c < f(b)$,
there is a $p \in (a,b)$ such that $f(p) =c$.
