If $f$ is a monic cubic polynomial with $f(0)=-64$, and all roots of $f$ are non-negative real numbers, what is the largest possible value of $f(-1)$? (A polynomial is monic if it has a leading coefficient of $1$.)
1. Given that \( f \) is a monic cubic polynomial with \( f(0) = -64 \) and all roots are non-negative real numbers, we can express \( f(x) \) as:
   \[
   f(x) = (x-a)(x-b)(x-c)
   \]
   where \( a, b, c \geq 0 \) and \( abc = -f(0) = 64 \).

2. We need to find the largest possible value of \( f(-1) \). First, we substitute \( x = -1 \) into the polynomial:
   \[
   f(-1) = (-1-a)(-1-b)(-1-c)
   \]
   Simplifying, we get:
   \[
   f(-1) = -(1+a)(1+b)(1+c)
   \]

3. To maximize \( f(-1) \), we need to minimize \( (1+a)(1+b)(1+c) \). Using the AM-GM inequality, we know:
   \[
   1 + a \geq 1, \quad 1 + b \geq 1, \quad 1 + c \geq 1
   \]
   Therefore:
   \[
   (1+a)(1+b)(1+c) \geq (1 + \sqrt[3]{abc})^3
   \]
   Since \( abc = 64 \), we have:
   \[
   \sqrt[3]{abc} = \sqrt[3]{64} = 4
   \]
   Thus:
   \[
   (1 + 4)^3 = 5^3 = 125
   \]

4. The equality in the AM-GM inequality holds when \( 1 + a = 1 + b = 1 + c \), which implies \( a = b = c = 4 \). Therefore, the polynomial \( f(x) \) can be written as:
   \[
   f(x) = (x-4)^3
   \]

5. Substituting \( x = -1 \) into this polynomial, we get:
   \[
   f(-1) = ( -1 - 4 )^3 = (-5)^3 = -125
   \]

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

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