Let $f$ be a monic cubic polynomial such that the sum of the coefficients of $f$ is $5$ and such that the sum of the roots of $f$ is $1$. Find the absolute value of the sum of the cubes of the roots of $f$.
1. Let \( f(x) \) be a monic cubic polynomial. Since it is monic, it can be written as:
   \[
   f(x) = x^3 + ax^2 + bx + c
   \]
   Given that the sum of the coefficients of \( f \) is 5, we have:
   \[
   1 + a + b + c = 5 \implies a + b + c = 4
   \]

2. The sum of the roots of \( f \) is given as 1. By Vieta's formulas, the sum of the roots (taken one at a time) of the polynomial \( x^3 + ax^2 + bx + c \) is \(-a\). Therefore:
   \[
   -a = 1 \implies a = -1
   \]

3. Substituting \( a = -1 \) into the equation \( a + b + c = 4 \):
   \[
   -1 + b + c = 4 \implies b + c = 5
   \]

4. Let the roots of \( f(x) \) be \( r_1, r_2, r_3 \). By Vieta's formulas, we have:
   \[
   r_1 + r_2 + r_3 = -a = 1
   \]
   \[
   r_1r_2 + r_2r_3 + r_3r_1 = b
   \]
   \[
   r_1r_2r_3 = -c
   \]

5. We need to find the absolute value of the sum of the cubes of the roots, \( |r_1^3 + r_2^3 + r_3^3| \). Using the identity for the sum of cubes:
   \[
   r_1^3 + r_2^3 + r_3^3 = (r_1 + r_2 + r_3)(r_1^2 + r_2^2 + r_3^2 - r_1r_2 - r_2r_3 - r_3r_1) + 3r_1r_2r_3
   \]

6. First, calculate \( r_1^2 + r_2^2 + r_3^2 \) using the square of the sum of the roots:
   \[
   (r_1 + r_2 + r_3)^2 = r_1^2 + r_2^2 + r_3^2 + 2(r_1r_2 + r_2r_3 + r_3r_1)
   \]
   \[
   1^2 = r_1^2 + r_2^2 + r_3^2 + 2b
   \]
   \[
   1 = r_1^2 + r_2^2 + r_3^2 + 2b \implies r_1^2 + r_2^2 + r_3^2 = 1 - 2b
   \]

7. Substitute \( r_1 + r_2 + r_3 = 1 \), \( r_1^2 + r_2^2 + r_3^2 = 1 - 2b \), and \( r_1r_2r_3 = -c \) into the sum of cubes identity:
   \[
   r_1^3 + r_2^3 + r_3^3 = 1 \cdot (1 - 2b - b) + 3(-c)
   \]
   \[
   r_1^3 + r_2^3 + r_3^3 = 1 \cdot (1 - 3b) - 3c
   \]
   \[
   r_1^3 + r_2^3 + r_3^3 = 1 - 3b - 3c
   \]

8. From \( b + c = 5 \), we can express \( c \) as \( c = 5 - b \):
   \[
   r_1^3 + r_2^3 + r_3^3 = 1 - 3b - 3(5 - b)
   \]
   \[
   r_1^3 + r_2^3 + r_3^3 = 1 - 3b - 15 + 3b
   \]
   \[
   r_1^3 + r_2^3 + r_3^3 = 1 - 15
   \]
   \[
   r_1^3 + r_2^3 + r_3^3 = -14
   \]

9. The absolute value of the sum of the cubes of the roots is:
   \[
   |r_1^3 + r_2^3 + r_3^3| = |-14| = 14
   \]

The final answer is \(\boxed{14}\).