Find the number of real numbers that are not in the domain of the function $$f(x) = \frac{1}{x} + \frac{1}{x^2} + \frac{1}{x^3}.$$

Answer 1

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.