Find all integers $k$ such that all roots of the following polynomial are also integers: $$f(x)=x^3-(k-3)x^2-11x+(4k-8).$$
To find all integers \( k \) such that all roots of the polynomial \( f(x) = x^3 - (k-3)x^2 - 11x + (4k-8) \) are also integers, we can proceed as follows:

1. **Check for integer roots using the Rational Root Theorem:**
   The Rational Root Theorem states that any rational root of the polynomial \( f(x) \) must be a factor of the constant term divided by a factor of the leading coefficient. Here, the constant term is \( 4k - 8 \) and the leading coefficient is 1. Therefore, any potential rational root must be an integer factor of \( 4k - 8 \).

2. **Evaluate \( f(x) \) at potential integer roots:**
   We need to check the values of \( f(x) \) at potential integer roots. Let's start by evaluating \( f(x) \) at \( x = 2 \) and \( x = -2 \):
   \[
   f(2) = 2^3 - (k-3)2^2 - 11 \cdot 2 + (4k-8) = 8 - 4(k-3) - 22 + 4k - 8
   \]
   Simplifying, we get:
   \[
   f(2) = 8 - 4k + 12 - 22 + 4k - 8 = -10
   \]
   \[
   f(-2) = (-2)^3 - (k-3)(-2)^2 - 11(-2) + (4k-8) = -8 - 4(k-3) + 22 + 4k - 8
   \]
   Simplifying, we get:
   \[
   f(-2) = -8 - 4k + 12 + 22 + 4k - 8 = 18
   \]

3. **Apply the Intermediate Value Theorem:**
   Since \( f(2) = -10 \) and \( f(-2) = 18 \), and the polynomial is continuous, by the Intermediate Value Theorem, there must be a root between \( -2 \) and \( 2 \). Since we are looking for integer roots, we need to check the values \( -1, 0, 1 \).

4. **Check integer roots \( -1, 0, 1 \):**
   \[
   f(-1) = (-1)^3 - (k-3)(-1)^2 - 11(-1) + (4k-8) = -1 - (k-3) + 11 + 4k - 8
   \]
   Simplifying, we get:
   \[
   f(-1) = -1 - k + 3 + 11 + 4k - 8 = 3k + 5
   \]
   Setting \( f(-1) = 0 \):
   \[
   3k + 5 = 0 \implies k = -\frac{5}{3}
   \]
   Since \( k \) must be an integer, \( k = -\frac{5}{3} \) is not valid.

   \[
   f(0) = 0^3 - (k-3)0^2 - 11 \cdot 0 + (4k-8) = 4k - 8
   \]
   Setting \( f(0) = 0 \):
   \[
   4k - 8 = 0 \implies k = 2
   \]
   \[
   f(1) = 1^3 - (k-3)1^2 - 11 \cdot 1 + (4k-8) = 1 - (k-3) - 11 + 4k - 8
   \]
   Simplifying, we get:
   \[
   f(1) = 1 - k + 3 - 11 + 4k - 8 = 3k - 15
   \]
   Setting \( f(1) = 0 \):
   \[
   3k - 15 = 0 \implies k = 5
   \]

5. **Verify the roots for \( k = 5 \):**
   For \( k = 5 \):
   \[
   f(x) = x^3 - (5-3)x^2 - 11x + (4 \cdot 5 - 8) = x^3 - 2x^2 - 11x + 12
   \]
   We need to check if this polynomial has integer roots. Using the Rational Root Theorem, the potential roots are \( \pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12 \).

   Checking \( x = 1 \):
   \[
   f(1) = 1^3 - 2 \cdot 1^2 - 11 \cdot 1 + 12 = 1 - 2 - 11 + 12 = 0
   \]
   So, \( x = 1 \) is a root. We can factor \( f(x) \) as:
   \[
   f(x) = (x-1)(x^2 - x - 12)
   \]
   Solving \( x^2 - x - 12 = 0 \):
   \[
   x^2 - x - 12 = (x-4)(x+3)
   \]
   So, the roots are \( x = 1, 4, -3 \), which are all integers.

The final answer is \( \boxed{ k = 5 } \)