A cubic polynomial P is such that P(1) = 1, P(2) = 2, P(3) = 3 and P(4) = 5. Then P(6) is

Answer

To solve the problem, we need to find the value of the cubic polynomial \( P(x) \) given the conditions \( P(1) = 1 \), \( P(2) = 2 \), \( P(3) = 3 \), and \( P(4) = 5 \).

### Step-by-step Solution:

1. **Define the Polynomial**:
Since \( P(x) \) is a cubic polynomial, we can express it as:
\[ P(x) = ax^3 + bx^2 + cx + d \]

2. **Construct a New Function**:
We can define a new function \( f(x) = P(x) - x \). This function will have roots at \( x = 1, 2, 3 \) because:
\[ f(1) = P(1) - 1 = 0, \quad f(2) = P(2) - 2 = 0, \quad f(3) = P(3) - 3 = 0 \]
Therefore, we can express \( f(x) \) as:
\[ f(x) = a(x - 1)(x - 2)(x - 3) \]

3. **Relate \( P(x) \) to \( f(x) \)**:
Since \( P(x) = f(x) + x \), we have:
\[ P(x) = a(x - 1)(x - 2)(x - 3) + x \]

4. **Use the Given Condition**:
We also know that \( P(4) = 5 \). Substituting \( x = 4 \) into our expression for \( P(x) \):
\[ P(4) = a(4 - 1)(4 - 2)(4 - 3) + 4 = 5 \]
Simplifying this gives:
\[ P(4) = a(3)(2)(1) + 4 = 5 \]
\[ 6a + 4 = 5 \]
\[ 6a = 1 \quad \Rightarrow \quad a = \frac{1}{6} \]

5. **Substitute \( a \) Back into \( P(x) \)**:
Now we can write \( P(x) \):
\[ P(x) = \frac{1}{6}(x - 1)(x - 2)(x - 3) + x \]

6. **Find \( P(6) \)**:
Now, we substitute \( x = 6 \) into \( P(x) \):
\[ P(6) = \frac{1}{6}(6 - 1)(6 - 2)(6 - 3) + 6 \]
Simplifying this gives:
\[ P(6) = \frac{1}{6}(5)(4)(3) + 6 \]
\[ = \frac{1}{6}(60) + 6 \]
\[ = 10 + 6 = 16 \]

### Final Answer:
Thus, the value of \( P(6) \) is:
\[ \boxed{16} \]