Checking whether a function is even or odd and checking if a function is periodic For given function, for example $f(x)=x^3+x^2-x-1$, to check whether it's even or odd, we have to find $f(-x)$.
Therefore, $f(-x)=-x^3+x^2+x-1$, which proves the function is not odd neither even.
Now, once we have that, how do I check whether the function is periodic?
 A: Where is $f(x) = x^3+x^2-x-1 = 0\,$?
E.g., Hint: $f(1)=0.$
And note $f(x) = (x - 1)(x^2 + 2x + 1) = (x-1)(x+1)^2$. So the only other "zero" of the function, is at $x = -1$: $\quad f(-1) = 0$.
A periodic function with period $p$ (as suggested in the comments: it doesn't matter what the value of $p$ is), has infinitely $x$ such that $f(x)=0,$ and it this case, $f(1) = 0$, there must exist some $p$ such that $ x=1,1\pm p,1\pm 2p, \cdots$.
As Dilip asked above: "Is it possible for $f(x)=x^3+x^2−x−1$ to have the value $0$ for infinitely many values of $x$? Why or why not?"
Try graphing $f(x)$ for to develop some intuition about your function; doing so will make the behavior of $f(x)$ immediately apparent. Below, I've included graphs produced by WolframAlpha. As the graphs confirms, there are exactly two values of $x$ at which $f(x) = 0$. So what can you conclude with respect to whether or not $f(x)$ is periodic.
Close up view of function:

Wider range:

As the top graph confirms, there are exactly two values of $x$ at which $f(x) = 0$.  And you can see from the graphs that
$f(x)$ is increasing on $x \in (-\infty, -1) \cup (1, \infty)$, and it is decreasing only when $x \in (-1, 1)$.
Since $f(x)$ it has exactly two zeros, hence finitely many zeros, $f(x)$ cannot be periodic.
A: Hint: A periodic function can be monotonous?
A: If you know about polynomial end-behavior, and about its continuity (so boundedness on any closed interval), then you can conclude that $f$ is not periodic.
If not, you can take an arbitrary $P$ and suppose that $f(x+P)=f(x)$ for all $x$. Expand the terms of $$f(x+P)-f(x)=0,$$ and you'll end up with a polynomial in $x$ and $P$ with a factor of $P$. For such a polynomial to be $0$ for all $x$, we require $P$ to be $0$, so $f$ isn't periodic.
More simply, note that $$f(x)=x^2(x+1)-(x+1)=(x^2-1)(x+1)=(x-1)(x+1)^2,$$ so $f$ has $2$ real zeroes. If $f$ were periodic, then it would necessarily have infinitely-many zeroes.
