Let f(x) be a real valued periodic function with domain R such that
`f(x+p)=1+[2-3f(x)+3(f(x))^(2)-(f(x))^(3)]^(1//3)` hold good for all ` x in R ` and some positive constant p, then the periodic of f(x) is

Answer

To solve the problem, we need to analyze the given functional equation for the periodic function \( f(x) \):

\[ f(x + p) = 1 + [2 - 3f(x) + 3(f(x))^2 - (f(x))^3]^{1/3} \]

### Step 1: Rewrite the equation
We can rewrite the equation by letting \( g(x) = f(x) - 1 \). Thus, we have:

\[ f(x) = g(x) + 1 \]

Substituting this into the original equation gives:

\[ f(x + p) = 1 + [2 - 3(g(x) + 1) + 3(g(x) + 1)^2 - (g(x) + 1)^3]^{1/3} \]

### Step 2: Simplify the expression
Now, simplify the expression inside the brackets:

1. Expand \( (g(x) + 1)^2 \) and \( (g(x) + 1)^3 \):
- \( (g(x) + 1)^2 = g(x)^2 + 2g(x) + 1 \)
- \( (g(x) + 1)^3 = g(x)^3 + 3g(x)^2 + 3g(x) + 1 \)

2. Substitute these into the equation:
\[ f(x + p) = 1 + [2 - 3g(x) - 3 + 3(g(x)^2 + 2g(x) + 1) - (g(x)^3 + 3g(x)^2 + 3g(x) + 1)]^{1/3} \]

3. Combine like terms:
\[ = 1 + [2 - 3g(x) - 3 + 3g(x)^2 + 6g(x) + 3 - g(x)^3 - 3g(x)^2 - 3g(x) - 1]^{1/3} \]
\[ = 1 + [1 + 0g(x)^2 + 0g(x) + 0]^{1/3} \]
\[ = 1 + 1 = 2 \]

### Step 3: Establish periodicity
Now we have:

\[ f(x + p) = 2 \]

Next, we can check \( f(x + 2p) \):

Substituting \( f(x + p) = 2 \) back into the original equation, we find:

\[ f(x + 2p) = 1 + [2 - 3(2) + 3(2)^2 - (2)^3]^{1/3} \]
\[ = 1 + [2 - 6 + 12 - 8]^{1/3} \]
\[ = 1 + [0]^{1/3} = 1 \]

### Step 4: Verify periodicity
Now we can see that:

1. \( f(x) = 1 \)
2. \( f(x + p) = 2 \)
3. \( f(x + 2p) = 1 \)

This indicates that \( f(x) \) is periodic with a period of \( 2p \).

### Conclusion
Thus, the period of the function \( f(x) \) is:

\[ \boxed{2p} \]