How many real roots does this polynomial have?

Let $f(x)=x^3-x^2-1$

Taking the first derivateve and setting it equal to $0$ to find the critical points:

$f'(x)=3x^2-2x=0$

$x(3x-2)=0 \quad \iff \quad x=0 \vee x=\dfrac{2}{3}$

Taking the second derivative and evaluating it at the critical points to check which is the local maximum and which is the local minimum:

$f''(0)=6\cdot 0-2<0 \Rightarrow local\phantom{c}maximum$

$f''\left(\dfrac{2}{3}\right)=6\cdot\dfrac{2}{3} -2>0 \Rightarrow local\phantom{c}minimum$

Since $f(0)=-1<0$ , $f(x)$ has only $\boxed{1}$ real root

This is a polynomial of degree 3, therefore, it will have three roots in total. To determine the number of real roots, we make individual plots of the functions $x^3$ and $x^2+1$ . The point(s) at which the graphs of these two functions intersect will be the roots of the given polynomial. The individual functions are plotted below:

From the figure, we see that there is only one point of intersection. Therefore, this polynomial has only one real root and the remaining two are complex.