Let $f(x)$ be a monic cubic polynomial with $f(0)=-64$, and all roots of $f(x)$ are non-negative real numbers. What is the largest possible value of $f(-1)$? (A polynomial is monic if its leading coefficient is 1.)
1. Let the roots of the monic cubic polynomial \( f(x) \) be \( r_1, r_2, r_3 \). Since \( f(x) \) is monic, we can write it as:
   \[
   f(x) = (x - r_1)(x - r_2)(x - r_3)
   \]
2. Given that \( f(0) = -64 \), we substitute \( x = 0 \) into the polynomial:
   \[
   f(0) = (-r_1)(-r_2)(-r_3) = -r_1 r_2 r_3 = -64
   \]
   Therefore, we have:
   \[
   r_1 r_2 r_3 = 64
   \]
3. We need to find the maximum possible value of \( f(-1) \). Substituting \( x = -1 \) into the polynomial, we get:
   \[
   f(-1) = (-1 - r_1)(-1 - r_2)(-1 - r_3) = (1 + r_1)(1 + r_2)(1 + r_3)
   \]
4. To maximize \( f(-1) \), we need to maximize \( (1 + r_1)(1 + r_2)(1 + r_3) \). We use the AM-GM inequality to find the maximum value:
   \[
   (1 + r_1)(1 + r_2)(1 + r_3) \geq \left( \sqrt[3]{(1 + r_1)(1 + r_2)(1 + r_3)} \right)^3
   \]
5. By Hölder's inequality, we have:
   \[
   (1 + r_1)(1 + r_2)(1 + r_3) \geq \left( 1^{\frac{1}{3}} + (r_1 r_2 r_3)^{\frac{1}{3}} \right)^3
   \]
6. Since \( r_1 r_2 r_3 = 64 \), we get:
   \[
   (1 + r_1)(1 + r_2)(1 + r_3) \geq \left( 1 + 4 \right)^3 = 5^3 = 125
   \]
7. The equality case in Hölder's inequality is achieved when \( r_1 = r_2 = r_3 = 4 \). Substituting these values, we get:
   \[
   f(x) = (x - 4)^3
   \]
   \[
   f(-1) = (-1 - 4)^3 = (-5)^3 = -125
   \]

Thus, the largest possible value of \( f(-1) \) is \( -125 \).

The final answer is \( \boxed{-125} \).