Let a,b,c, d be the roots of `x ^(4) -x ^(3)-x ^(2) -1=0.` Also consider `P (x) =x ^(6)-x ^(5) -x ^(3) -x ^(2) -x,` then the value of `p (a) +p(b) +p(c ) +p(d)` is equal to :

Answer

To solve the problem, we need to find the value of \( P(a) + P(b) + P(c) + P(d) \) where \( a, b, c, d \) are the roots of the polynomial \( x^4 - x^3 - x^2 - 1 = 0 \) and \( P(x) = x^6 - x^5 - x^3 - x^2 - x \).

### Step 1: Identify the polynomial \( Q(x) \)
We start with the polynomial \( Q(x) = x^4 - x^3 - x^2 - 1 \). Since \( a, b, c, d \) are the roots of \( Q(x) \), we know that \( Q(a) = Q(b) = Q(c) = Q(d) = 0 \).

### Step 2: Express \( P(x) \) in terms of \( Q(x) \)
We can express \( P(x) \) in terms of \( Q(x) \). We can perform polynomial long division of \( P(x) \) by \( Q(x) \):

\[ P(x) = x^6 - x^5 - x^3 - x^2 - x \]

Dividing \( P(x) \) by \( Q(x) \):

1. Divide \( x^6 \) by \( x^4 \) to get \( x^2 \).
2. Multiply \( Q(x) \) by \( x^2 \) to get \( x^6 - x^5 - x^4 - x^2 \).
3. Subtract this from \( P(x) \):

\[ P(x) - (x^6 - x^5 - x^4 - x^2) = x^4 - x^3 - x \]

4. Now, divide \( x^4 - x^3 - x \) by \( Q(x) \):

Since \( x^4 - x^3 - x \) is of lower degree than \( Q(x) \), we can express:

\[ P(x) = Q(x) \cdot (x^2) + (x^4 - x^3 - x) \]

### Step 3: Evaluate \( P(a) \)
Since \( Q(a) = 0 \), we have:

\[ P(a) = 0 \cdot (a^2) + (a^4 - a^3 - a) = a^4 - a^3 - a \]

Similarly, we can find:

\[ P(b) = b^4 - b^3 - b \]
\[ P(c) = c^4 - c^3 - c \]
\[ P(d) = d^4 - d^3 - d \]

### Step 4: Sum \( P(a) + P(b) + P(c) + P(d) \)
Now we can sum these:

\[ P(a) + P(b) + P(c) + P(d) = (a^4 - a^3 - a) + (b^4 - b^3 - b) + (c^4 - c^3 - c) + (d^4 - d^3 - d) \]

This simplifies to:

\[ = (a^4 + b^4 + c^4 + d^4) - (a^3 + b^3 + c^3 + d^3) - (a + b + c + d) \]

### Step 5: Use Vieta's Formulas
Using Vieta's formulas for the roots of \( Q(x) \):

- \( a + b + c + d = 1 \)
- \( ab + ac + ad + bc + bd + cd = -1 \)
- \( abc + abd + acd + bcd = 0 \)
- \( abcd = 1 \)

To find \( a^2 + b^2 + c^2 + d^2 \):

\[ a^2 + b^2 + c^2 + d^2 = (a + b + c + d)^2 - 2(ab + ac + ad + bc + bd + cd) = 1^2 - 2(-1) = 1 + 2 = 3 \]

Now, we can find \( a^3 + b^3 + c^3 + d^3 \):

\[ a^3 + b^3 + c^3 + d^3 = (a + b + c + d)(a^2 + b^2 + c^2 + d^2 - (ab + ac + ad + bc + bd + cd)) = 1(3 - (-1)) = 1 \cdot 4 = 4 \]

Now we can find \( a^4 + b^4 + c^4 + d^4 \):

Using the identity \( a^4 + b^4 + c^4 + d^4 = (a^2 + b^2 + c^2 + d^2)^2 - 2(a^2b^2 + a^2c^2 + a^2d^2 + b^2c^2 + b^2d^2 + c^2d^2) \).

To find \( a^2b^2 + a^2c^2 + a^2d^2 + b^2c^2 + b^2d^2 + c^2d^2 \):

Using \( (ab + ac + ad + bc + bd + cd)^2 = a^2b^2 + a^2c^2 + a^2d^2 + b^2c^2 + b^2d^2 + c^2d^2 + 2(abc + abd + acd + bcd) \):

\[ (-1)^2 = 1 = a^2b^2 + a^2c^2 + a^2d^2 + b^2c^2 + b^2d^2 + c^2d^2 + 0 \]
\[ \Rightarrow a^2b^2 + a^2c^2 + a^2d^2 + b^2c^2 + b^2d^2 + c^2d^2 = 1 \]

Now substituting back:

\[ a^4 + b^4 + c^4 + d^4 = 3^2 - 2(1) = 9 - 2 = 7 \]

### Final Calculation
Now substituting back into the sum:

\[ P(a) + P(b) + P(c) + P(d) = (7) - (4) - (1) = 7 - 4 - 1 = 2 \]

### Conclusion
Thus, the value of \( P(a) + P(b) + P(c) + P(d) \) is:

\[ \boxed{2} \]