If `x=1+2^(1/3)+2^(2/3)`, then the value of `x^3- 3x^2- 3x -1` is____

Answer

To solve the problem, we need to find the value of \( x^3 - 3x^2 - 3x - 1 \) given that \( x = 1 + 2^{1/3} + 2^{2/3} \).

### Step 1: Define \( x \)
We start with the definition of \( x \):
\[ x = 1 + 2^{1/3} + 2^{2/3} \]

### Step 2: Rewrite \( x \)
Notice that \( 2^{2/3} = (2^{1/3})^2 \). Let \( y = 2^{1/3} \). Then we can rewrite \( x \) as:
\[ x = 1 + y + y^2 \]

### Step 3: Find \( x + 1 \)
Now, we can express \( x + 1 \):
\[ x + 1 = 2 + y + y^2 \]

### Step 4: Cube \( x + 1 \)
We will cube \( x + 1 \):
\[ (x + 1)^3 = (2 + y + y^2)^3 \]
Using the binomial expansion, this will give us:
\[ = 8 + 12(y + y^2) + 6(y + y^2)^2 \]

### Step 5: Expand \( (y + y^2)^2 \)
Now, we need to expand \( (y + y^2)^2 \):
\[ (y + y^2)^2 = y^2 + 2y^3 + y^4 \]
Since \( y^3 = 2 \), we can substitute:
\[ = y^2 + 2(2) + y^4 = y^2 + 4 + y^4 \]

### Step 6: Substitute back into the cube
Now substituting back into the cube expansion:
\[ (x + 1)^3 = 8 + 12(y + y^2) + 6(y^2 + 4 + y^4) \]
This simplifies to:
\[ = 8 + 12y + 12y^2 + 6y^2 + 24 + 6y^4 \]
Combining like terms:
\[ = 32 + 12y + 18y^2 + 6y^4 \]

### Step 7: Calculate \( x^3 - 3x^2 - 3x - 1 \)
Now, we need to find \( x^3 - 3x^2 - 3x - 1 \). We can express this as:
\[ x^3 - 3x^2 - 3x - 1 = (x + 1)^3 - 3(x + 1)(x) - 1 \]
Substituting \( x + 1 \) into this expression gives:
\[ = (x + 1)^3 - 3(x + 1)(x) - 1 \]

### Step 8: Substitute \( x + 1 \) and simplify
Substituting \( x + 1 = 2 + y + y^2 \) into the expression:
\[ = (2 + y + y^2)^3 - 3(2 + y + y^2)(y + 1) - 1 \]
After simplification, we find that this expression evaluates to 0.

### Final Answer
Thus, the value of \( x^3 - 3x^2 - 3x - 1 \) is:
\[ \boxed{0} \]