Are there positive integers $a; b; c$ and $d$ such that $a^3 + b^3 + c^3 + d^3 =100^{100}$ ?

[i](4 points)[/i]
To determine if there are positive integers \(a, b, c,\) and \(d\) such that \(a^3 + b^3 + c^3 + d^3 = 100^{100}\), we can proceed as follows:

1. **Simplify the problem by considering the magnitude of the numbers:**
   \[
   100^{100} = (10^2)^{100} = 10^{200}
   \]
   We need to find positive integers \(a, b, c,\) and \(d\) such that:
   \[
   a^3 + b^3 + c^3 + d^3 = 10^{200}
   \]

2. **Assume a common factor for simplicity:**
   Let \(a, b, c,\) and \(d\) be multiples of \(10^{66}\). That is, let:
   \[
   a = 10^{66}a_1, \quad b = 10^{66}b_1, \quad c = 10^{66}c_1, \quad d = 10^{66}d_1
   \]
   Then:
   \[
   a^3 = (10^{66}a_1)^3 = 10^{198}a_1^3
   \]
   Similarly:
   \[
   b^3 = 10^{198}b_1^3, \quad c^3 = 10^{198}c_1^3, \quad d^3 = 10^{198}d_1^3
   \]
   Therefore, the equation becomes:
   \[
   10^{198}(a_1^3 + b_1^3 + c_1^3 + d_1^3) = 10^{200}
   \]
   Dividing both sides by \(10^{198}\):
   \[
   a_1^3 + b_1^3 + c_1^3 + d_1^3 = 10^2 = 100
   \]

3. **Find integers \(a_1, b_1, c_1,\) and \(d_1\) such that their cubes sum to 100:**
   We need to solve:
   \[
   a_1^3 + b_1^3 + c_1^3 + d_1^3 = 100
   \]
   One possible solution is:
   \[
   1^3 + 2^3 + 3^3 + 4^3 = 1 + 8 + 27 + 64 = 100
   \]
   Thus, \(a_1, b_1, c_1,\) and \(d_1\) can be \(1, 2, 3,\) and \(4\) in some permutation.

4. **Construct the original numbers \(a, b, c,\) and \(d\):**
   Using the values of \(a_1, b_1, c_1,\) and \(d_1\), we get:
   \[
   a = 10^{66} \cdot 1 = 10^{66}, \quad b = 10^{66} \cdot 2 = 2 \cdot 10^{66}, \quad c = 10^{66} \cdot 3 = 3 \cdot 10^{66}, \quad d = 10^{66} \cdot 4 = 4 \cdot 10^{66}
   \]

5. **Verify the solution:**
   \[
   a^3 + b^3 + c^3 + d^3 = (10^{66})^3 + (2 \cdot 10^{66})^3 + (3 \cdot 10^{66})^3 + (4 \cdot 10^{66})^3
   \]
   \[
   = 10^{198} + 8 \cdot 10^{198} + 27 \cdot 10^{198} + 64 \cdot 10^{198}
   \]
   \[
   = (1 + 8 + 27 + 64) \cdot 10^{198} = 100 \cdot 10^{198} = 10^{200}
   \]
   This confirms that the solution is correct.

The final answer is \( \boxed{ a = 10^{66}, b = 2 \cdot 10^{66}, c = 3 \cdot 10^{66}, d = 4 \cdot 10^{66} } \)