How many real numbers are not in the domain of the function $$f(x) = \frac{1}{x-64} + \frac{1}{x^2-64} + \frac{1}{x^3-64}~?$$
To find the domain of a function, I need to identify the values of x that would make the function undefined, such as dividing by zero or taking the square root of a negative number. In this case, the only issue is dividing by zero, since there are no square roots or other operations that could cause trouble. So, I need to avoid the values of x that would make any of the denominators equal to zero. To find those values, I can set each denominator equal to zero and solve for x. For the first denominator, I have $$x-64 = 0,$$ which implies that $$x = 64.$$ For the second denominator, I have $$x^2-64 = 0,$$ which is a difference of squares that can be factored as $$(x-8)(x+8) = 0.$$ This implies that $$x = 8$$ or $$x = -8.$$ For the third denominator, I have $$x^3-64 = 0,$$ which is a difference of cubes that can be factored as $$(x-4)(x^2+4x+16) = 0.$$ This implies that $$x = 4$$ or $$x^2+4x+16 = 0.$$ The latter equation is a quadratic that has no real solutions, since its discriminant is negative. Therefore, the only real values of x that are not in the domain of the function are 64, 8, -8, and 4. There are four such values, so that is the final answer. 