Let `p(x) =x^6-x^5-x^3-x^2-x` and `alpha, beta, gamma, delta` are the roots of the equation `x^4-x^3-x^2-1=0` then `P(alpha)+P(beta)+P(gamma)+P(delta)=`

Answer

To solve the problem, we need to evaluate \( P(\alpha) + P(\beta) + P(\gamma) + P(\delta) \) where \( P(x) = x^6 - x^5 - x^3 - x^2 - x \) and \( \alpha, \beta, \gamma, \delta \) are the roots of the equation \( x^4 - x^3 - x^2 - 1 = 0 \).

### Step 1: Express \( P(x) \)
We start with the polynomial:
\[ P(x) = x^6 - x^5 - x^3 - x^2 - x \]

### Step 2: Substitute the roots into \( P(x) \)
We will evaluate \( P(\alpha) \):
\[ P(\alpha) = \alpha^6 - \alpha^5 - \alpha^3 - \alpha^2 - \alpha \]

### Step 3: Use the polynomial equation
Since \( \alpha \) is a root of \( x^4 - x^3 - x^2 - 1 = 0 \), we can express higher powers of \( \alpha \) in terms of lower powers:
\[ \alpha^4 = \alpha^3 + \alpha^2 + 1 \]
Now we can find \( \alpha^5 \) and \( \alpha^6 \):
\[ \alpha^5 = \alpha \cdot \alpha^4 = \alpha(\alpha^3 + \alpha^2 + 1) = \alpha^4 + \alpha^3 + \alpha = (\alpha^3 + \alpha^2 + 1) + \alpha^3 + \alpha = 2\alpha^3 + \alpha^2 + \alpha + 1 \]
\[ \alpha^6 = \alpha \cdot \alpha^5 = \alpha(2\alpha^3 + \alpha^2 + \alpha + 1) = 2\alpha^4 + \alpha^3 + \alpha^2 + \alpha = 2(\alpha^3 + \alpha^2 + 1) + \alpha^3 + \alpha^2 + \alpha \]
\[ = 3\alpha^3 + 3\alpha^2 + 2 + \alpha \]

### Step 4: Substitute \( \alpha^5 \) and \( \alpha^6 \) back into \( P(\alpha) \)
Now substituting these back into \( P(\alpha) \):
\[ P(\alpha) = (3\alpha^3 + 3\alpha^2 + 2 + \alpha) - (2\alpha^3 + \alpha^2 + \alpha + 1) - \alpha^3 - \alpha^2 - \alpha \]
Simplifying this:
\[ = 3\alpha^3 + 3\alpha^2 + 2 + \alpha - 2\alpha^3 - \alpha^2 - \alpha - 1 - \alpha^3 - \alpha^2 - \alpha \]
\[ = (3\alpha^3 - 2\alpha^3 - \alpha^3) + (3\alpha^2 - \alpha^2 - \alpha^2) + (2 - 1) + (\alpha - \alpha - \alpha) \]
\[ = 0\alpha^3 + 1\alpha^2 + 1 + 0\alpha = \alpha^2 + 1 \]

### Step 5: Calculate \( P(\alpha) + P(\beta) + P(\gamma) + P(\delta) \)
Now we can sum up \( P(\alpha) + P(\beta) + P(\gamma) + P(\delta) \):
\[ P(\alpha) + P(\beta) + P(\gamma) + P(\delta) = (\alpha^2 + 1) + (\beta^2 + 1) + (\gamma^2 + 1) + (\delta^2 + 1) \]
\[ = (\alpha^2 + \beta^2 + \gamma^2 + \delta^2) + 4 \]

### Step 6: Use the identity for the sum of squares
Using the identity:
\[ \alpha^2 + \beta^2 + \gamma^2 + \delta^2 = (\alpha + \beta + \gamma + \delta)^2 - 2(\alpha\beta + \alpha\gamma + \alpha\delta + \beta\gamma + \beta\delta + \gamma\delta) \]
From the polynomial \( x^4 - x^3 - x^2 - 1 = 0 \):
- The sum of the roots \( \alpha + \beta + \gamma + \delta = 1 \)
- The sum of the products of the roots taken two at a time \( \alpha\beta + \alpha\gamma + \alpha\delta + \beta\gamma + \beta\delta + \gamma\delta = -1 \)

Thus:
\[ \alpha^2 + \beta^2 + \gamma^2 + \delta^2 = 1^2 - 2(-1) = 1 + 2 = 3 \]

### Final Calculation
Now substituting back:
\[ P(\alpha) + P(\beta) + P(\gamma) + P(\delta) = 3 + 4 = 7 \]

### Conclusion
The final answer is:
\[ \boxed{7} \]