Find all numbers between $100$ and $999$ which equal the sum of the cubes of their digits.
To find all numbers between \(100\) and \(999\) which equal the sum of the cubes of their digits, we will denote our 3-digit number as \(\overline{abc}\), where \(a\), \(b\), and \(c\) are the digits of the number. This means the number can be expressed as \(100a + 10b + c\).

We need to find all such numbers that satisfy the equation:
\[ 100a + 10b + c = a^3 + b^3 + c^3 \]

1. **Analyzing the equation modulo 10:**
   \[
   100a + 10b + c \equiv a^3 + b^3 + c^3 \pmod{10}
   \]
   Since \(100a \equiv 0 \pmod{10}\) and \(10b \equiv 0 \pmod{10}\), we have:
   \[
   c \equiv a^3 + b^3 + c^3 \pmod{10}
   \]
   Simplifying, we get:
   \[
   a^3 + b^3 \equiv c - c^3 \pmod{10}
   \]

2. **Checking possible values of \(a\):**
   We will check each possible value of \(a\) from 1 to 9 (since \(a\) cannot be 0 in a 3-digit number).

   - **For \(a = 1\):**
     \[
     1^3 = 1 \implies 1 - 1^3 \equiv 0 \pmod{10}
     \]
     \[
     b^3 + c^3 \equiv 0 \pmod{10}
     \]
     Possible pairs \((b, c)\) are \((1, 9), (2, 8), (3, 7), (4, 6), (5, 5), (6, 4), (7, 3), (8, 2), (9, 1)\).

   - **For \(a = 2\):**
     \[
     2^3 = 8 \implies 2 - 2^3 \equiv 4 \pmod{10}
     \]
     \[
     b^3 + c^3 \equiv 4 \pmod{10}
     \]
     Possible pairs \((b, c)\) are \((0, 4), (1, 7), (2, 6), (3, 3), (4, 9), (5, 2), (6, 1), (7, 0), (8, 8), (9, 5)\).

   - **For \(a = 3\):**
     \[
     3^3 = 27 \implies 3 - 3^3 \equiv 6 \pmod{10}
     \]
     \[
     b^3 + c^3 \equiv 6 \pmod{10}
     \]
     Possible pairs \((b, c)\) are \((0, 6), (1, 5), (2, 2), (3, 9), (4, 8), (5, 1), (6, 0), (7, 7), (8, 4), (9, 5)\).

   - **For \(a = 4\):**
     \[
     4^3 = 64 \implies 4 - 4^3 \equiv 0 \pmod{10}
     \]
     \[
     b^3 + c^3 \equiv 0 \pmod{10}
     \]
     Possible pairs \((b, c)\) are \((1, 9), (2, 8), (3, 7), (4, 6), (5, 5), (6, 4), (7, 3), (8, 2), (9, 1)\).

   - **For \(a = 5\):**
     \[
     5^3 = 125 \implies 5 - 5^3 \equiv 0 \pmod{10}
     \]
     \[
     b^3 + c^3 \equiv 0 \pmod{10}
     \]
     Possible pairs \((b, c)\) are \((1, 9), (2, 8), (3, 7), (4, 6), (5, 5), (6, 4), (7, 3), (8, 2), (9, 1)\).

   - **For \(a = 6\):**
     \[
     6^3 = 216 \implies 6 - 6^3 \equiv 0 \pmod{10}
     \]
     \[
     b^3 + c^3 \equiv 0 \pmod{10}
     \]
     Possible pairs \((b, c)\) are \((1, 9), (2, 8), (3, 7), (4, 6), (5, 5), (6, 4), (7, 3), (8, 2), (9, 1)\).

   - **For \(a = 7\):**
     \[
     7^3 = 343 \implies 7 - 7^3 \equiv 4 \pmod{10}
     \]
     \[
     b^3 + c^3 \equiv 4 \pmod{10}
     \]
     Possible pairs \((b, c)\) are \((0, 4), (1, 7), (2, 6), (3, 3), (4, 9), (5, 2), (6, 1), (7, 0), (8, 8), (9, 5)\).

   - **For \(a = 8\):**
     \[
     8^3 = 512 \implies 8 - 8^3 \equiv 6 \pmod{10}
     \]
     \[
     b^3 + c^3 \equiv 6 \pmod{10}
     \]
     Possible pairs \((b, c)\) are \((0, 6), (1, 5), (2, 2), (3, 9), (4, 8), (5, 1), (6, 0), (7, 7), (8, 4), (9, 5)\).

   - **For \(a = 9\):**
     \[
     9^3 = 729 \implies 9 - 9^3 \equiv 0 \pmod{10}
     \]
     \[
     b^3 + c^3 \equiv 0 \pmod{10}
     \]
     Possible pairs \((b, c)\) are \((1, 9), (2, 8), (3, 7), (4, 6), (5, 5), (6, 4), (7, 3), (8, 2), (9, 1)\).

3. **Manually checking each pair:**
   We now manually check each pair of \((b, c)\) for each value of \(a\) to see if they satisfy the original equation \(100a + 10b + c = a^3 + b^3 + c^3\).

   - For \(a = 1\), checking pairs:
     \[
     100(1) + 10(5) + 3 = 1^3 + 5^3 + 3^3 \implies 153 = 1 + 125 + 27 \implies 153 = 153
     \]
     So, \(153\) is a solution.

   - For \(a = 3\), checking pairs:
     \[
     100(3) + 10(7) + 0 = 3^3 + 7^3 + 0^3 \implies 370 = 27 + 343 + 0 \implies 370 = 370
     \]
     So, \(370\) is a solution.

     \[
     100(3) + 10(7) + 1 = 3^3 + 7^3 + 1^3 \implies 371 = 27 + 343 + 1 \implies 371 = 371
     \]
     So, \(371\) is a solution.

   - For \(a = 4\), checking pairs:
     \[
     100(4) + 10(0) + 7 = 4^3 + 0^3 + 7^3 \implies 407 = 64 + 0 + 343 \implies 407 = 407
     \]
     So, \(407\) is a solution.

Thus, the numbers between \(100\) and \(999\) which equal the sum of the cubes of their digits are \(153\), \(370\), \(371\), and \(407\).

The final answer is \(\boxed{153, 370, 371, 407}\).