Find the number of real numbers that are not in the domain of the function $$f(x) = \frac{1}{x} + \frac{1}{x^2} + \frac{1}{x^3}.$$
The expression has a $x^2$ term in the denominator, so $x$ can't be equal to 0. It also has a $x^3$ term in the denominator, so $x$ can't be equal to 0. This means that $x$ must be nonzero. If $x=0$, it is obvious that $1/x+1/x^2+1/x^3$ cannot be a real number. Therefore, the only restriction we have on the domain of the expression is that $x \neq 0$. Therefore, the solution is $\boxed{\infty}$, or infinite, since there are infinitely many real numbers that do not equal 0.
To find the numbers that are not in the domain of the function, we need to identify values of $x$ that would make the function undefined.
The function $f(x)$ is defined as the sum of three terms: $\frac{1}{x}$, $\frac{1}{x^2}$, and $\frac{1}{x^3}$.
First, we note that division by zero is undefined, so any value of $x$ that makes any of the denominators ($x$, $x^2$, or $x^3$) equal to zero will not be in the domain of the function.
For $\frac{1}{x}$, $x$ cannot be equal to 0 ($x \neq 0$).
For $\frac{1}{x^2}$, $x^2$ cannot be equal to 0 ($x^2 \neq 0$). This means $x$ cannot be equal to 0.
For $\frac{1}{x^3}$, $x^3$ cannot be equal to 0 ($x^3 \neq 0$). Again, $x$ cannot be equal to 0.
In summary, the number 0 does not belong to the domain of $f(x)$ because it makes the function undefined. All other real numbers are in the domain of $f(x)$.
Therefore, there are infinitely many real numbers in the domain of $f(x)$, and only one number, 0, is not in its domain.
To find the numbers that are not in the domain of the function, we need to identify the values of $x$ that would cause the function to be undefined. In this case, we would have to look for the values of $x$ that would make the denominator equal to zero, since division by zero is undefined.
The function $f(x)$ has three terms in the denominator: $\frac{1}{x}$, $\frac{1}{x^2}$, and $\frac{1}{x^3}$. Let's examine each individually.
1. For $\frac{1}{x}$ to be defined, $x$ cannot be equal to zero. Dividing any number by zero is undefined. So, $x \neq 0$.
2. For $\frac{1}{x^2}$ to be defined, $x^2$ cannot be equal to zero. This means $x$ cannot be equal to zero as well since any number squared becomes zero. So, $x \neq 0$.
3. For $\frac{1}{x^3}$ to be defined, $x^3$ cannot be equal to zero. Therefore, $x$ must not be zero. So, $x \neq 0$.
Combining all the restrictions, we find that $x \neq 0$. So, the function is not defined at $x = 0$.
To find the number of real numbers that are not in the domain of the function, we count the infinite values in the set of real numbers and subtract the one value that is not in the domain. Hence, there are infinitely many real numbers that are not in the domain of the function.