Compute the number of ordered quadruples of complex numbers $(a,b,c,d)$ such that
\[ (ax+by)^3 + (cx+dy)^3 = x^3 + y^3 \]
holds for all complex numbers $x, y$.

[i]Proposed by Evan Chen[/i]
To solve the problem, we need to find the number of ordered quadruples \((a, b, c, d)\) of complex numbers such that the equation
\[ (ax + by)^3 + (cx + dy)^3 = x^3 + y^3 \]
holds for all complex numbers \(x\) and \(y\).

1. **Expand the given equation:**
   \[
   (ax + by)^3 + (cx + dy)^3 = x^3 + y^3
   \]
   Expanding both sides, we get:
   \[
   (ax + by)^3 = a^3x^3 + 3a^2bx^2y + 3ab^2xy^2 + b^3y^3
   \]
   \[
   (cx + dy)^3 = c^3x^3 + 3c^2d x^2y + 3cd^2xy^2 + d^3y^3
   \]
   Adding these, we have:
   \[
   a^3x^3 + 3a^2bx^2y + 3ab^2xy^2 + b^3y^3 + c^3x^3 + 3c^2dx^2y + 3cd^2xy^2 + d^3y^3 = x^3 + y^3
   \]

2. **Equate coefficients of like terms:**
   For the equation to hold for all \(x\) and \(y\), the coefficients of \(x^3\), \(x^2y\), \(xy^2\), and \(y^3\) on both sides must be equal. This gives us the following system of equations:
   \[
   a^3 + c^3 = 1 \quad \text{(coefficient of } x^3\text{)}
   \]
   \[
   b^3 + d^3 = 1 \quad \text{(coefficient of } y^3\text{)}
   \]
   \[
   3a^2b + 3c^2d = 0 \quad \text{(coefficient of } x^2y\text{)}
   \]
   \[
   3ab^2 + 3cd^2 = 0 \quad \text{(coefficient of } xy^2\text{)}
   \]

3. **Simplify the system:**
   From the equations \(3a^2b + 3c^2d = 0\) and \(3ab^2 + 3cd^2 = 0\), we can simplify to:
   \[
   a^2b = -c^2d
   \]
   \[
   ab^2 = -cd^2
   \]

4. **Case analysis:**
   - **Case 1: One of \(a, b, c, d\) is zero.**
     - If \(a = 0\), then \(c \neq 0\) (from \(a^3 + c^3 = 1\)) and thus \(d = 0\) (from \(a^2b = -c^2d\)) and \(b \neq 0\) (from \(b^3 + d^3 = 1\)).
     - Similarly, if \(b = 0\), then \(a, d \neq 0\) and \(c = 0\).
     - In these cases, we have \(a^3 = d^3 = 1\) and \(b^3 = c^3 = 1\). Each of these equations has 3 solutions (since \(x^3 = 1\) has three complex solutions: \(1, \omega, \omega^2\) where \(\omega\) is a primitive cube root of unity). Therefore, there are \(3^2 = 9\) solutions for each sub-case, giving a total of \(18\) solutions in this case.

   - **Case 2: None of \(a, b, c, d\) is zero.**
     - Dividing the equations \(a^2b = -c^2d\) and \(ab^2 = -cd^2\) gives:
       \[
       \frac{a}{b} = \frac{c}{d} = k \quad \text{for some nonzero } k
       \]
     - This implies \(a = kb\) and \(c = kd\). Substituting into \(a^3 + c^3 = 1\), we get:
       \[
       (kb)^3 + (kd)^3 = 1 \implies k^3(b^3 + d^3) = 1
       \]
     - From \(b^3 + d^3 = 1\), we have:
       \[
       k^3 \cdot 1 = 1 \implies k^3 = 1
       \]
     - Thus, \(k\) must be one of the cube roots of unity: \(1, \omega, \omega^2\). However, substituting \(k = 1, \omega, \omega^2\) into the equations \(a = kb\) and \(c = kd\) does not yield any new solutions beyond those already counted in Case 1.

Therefore, the total number of ordered quadruples \((a, b, c, d)\) is \(18\).

The final answer is \(\boxed{18}\).