EDIT
Recall the statement of the intermediate value theorem.
Theorem If $f(x)$ is a real-valued continuous function on the interval $[a,b]$, then given any $y \in [\min(f(a),f(b)), \max(f(a),f(b))]$, there exists $c \in [a,b]$ such that $f(c) = y$.
The theorem guarantees us that given any value $y$ in-between $f(a)$ and $f(b)$, the continuous function $f(x)$ takes the value $y$ for some point in the interval $[a,b]$.
Now lets get back to our problem. Look at the function $f(x) = \cos(x) - x$.
We have $f(0) = 1 > 0$.
We also have that $f(1) = \cos(1) - 1$. But $\cos(x) < 1$, $\forall x \neq2 n\pi$, where $n \in \mathbb{Z}$. Clearly, $1 \neq 2 n \pi$, where $n \in \mathbb{Z}$. Hence, we have that $\cos(1) < 1 \implies f(1) < 0$.
Hence, we have a continuous function $f(x) = \cos(x) - x$ on the interval $[0,1]$ with $f(0) = 1$ and $f(1) = \cos(1) - 1<0$. ($a=0$, $b=1$, $f(a) = 1$ and $f(b) = \cos(1) -1 < 0$).
Note that $0$ lies in the interval $[\cos(1)-1,1]$. Hence, from the intermediate value theorem, there exists a $c \in [0,1]$ such that $f(c) = 0$.
This means that $c$ is a root of the equation. Hence, we have proved that there exists a root in the interval $[0,1]$.
Hint for existence: if you let
$$f(x) := x^3 + 2x - \frac{1}{x},$$
note that $f(\frac{1}{4}) < 0, f(1) > 0$. Hence, ...
Hint for value: to calculate the zero, note that
$$x^3 + 2x - \frac{1}{x} = 0 \Leftrightarrow y^2 + 2y -1 =0,$$
if you define $y := x^2$.
Best Answer
Let $f(x) = x^3 + x + 1$. Then $f$ is a polynomial, so its continuous. Then we notice that
$$f(-1) = -1, \text{ and } f(0) = 1.$$
But then by the intermediate value theorem, this means that for all $y\in[-1,1]$ we have some $x\in [-1,0]$ such that $f(x) = y$. Therefore we have some $z$ such that $f(z) = 0$.
Since $f(z) = 0$ we know that $z^3 + z + 1 = 0$ and so the equation $$x^3 + x + 1 = 0$$ has a solution, namely $z$.