The positive numbers $a, b, c,d,e$ are such that the following identity hold for all real number $x$:
$(x + a)(x + b)(x + c) = x^3 + 3dx^2 + 3x + e^3$. Find the smallest value of $d$.
1. Given the identity:
   \[
   (x + a)(x + b)(x + c) = x^3 + 3dx^2 + 3x + e^3
   \]
   we can expand the left-hand side and compare coefficients with the right-hand side.

2. Expanding the left-hand side:
   \[
   (x + a)(x + b)(x + c) = x^3 + (a+b+c)x^2 + (ab+bc+ca)x + abc
   \]

3. By comparing coefficients with the right-hand side \(x^3 + 3dx^2 + 3x + e^3\), we get the following system of equations:
   \[
   a + b + c = 3d
   \]
   \[
   ab + bc + ca = 3
   \]
   \[
   abc = e^3
   \]

4. To find the smallest value of \(d\), we use the well-known inequality for positive numbers \(a, b, c\):
   \[
   (a + b + c)^2 \geq 3(ab + bc + ca)
   \]

5. Substituting the given expressions into the inequality:
   \[
   (3d)^2 \geq 3 \cdot 3
   \]
   \[
   9d^2 \geq 9
   \]
   \[
   d^2 \geq 1
   \]
   \[
   d \geq 1
   \]

6. Therefore, the smallest value of \(d\) that satisfies the inequality is:
   \[
   d = 1
   \]

The final answer is \(\boxed{1}\)