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 \( P(x) \) at \( x = 6 \), given the conditions \( P(1) = 1 \), \( P(2) = 2 \), \( P(3) = 3 \), and \( P(4) = 5 \).

### Step-by-Step Solution:

1. **Define a New Polynomial**:
Let's define a new polynomial \( f(x) = P(x) - x \). This polynomial will help us analyze the roots based on the given conditions.

2. **Evaluate the New Polynomial at Given Points**:
Using the conditions provided:
- \( f(1) = P(1) - 1 = 1 - 1 = 0 \)
- \( f(2) = P(2) - 2 = 2 - 2 = 0 \)
- \( f(3) = P(3) - 3 = 3 - 3 = 0 \)
- \( f(4) = P(4) - 4 = 5 - 4 = 1 \)

From the above evaluations, we see that \( f(x) \) has roots at \( x = 1, 2, 3 \).

3. **Express \( f(x) \)**:
Since \( f(x) \) is a cubic polynomial with roots at \( x = 1, 2, 3 \), we can express it as:
\[ f(x) = a(x - 1)(x - 2)(x - 3) \]
where \( a \) is a constant.

4. **Use the Condition at \( x = 4 \) to Find \( a \)**:
We know \( f(4) = P(4) - 4 = 1 \). Thus:
\[ f(4) = a(4 - 1)(4 - 2)(4 - 3) = a(3)(2)(1) = 6a \]
Setting this equal to 1 gives:
\[ 6a = 1 \implies a = \frac{1}{6} \]

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

6. **Find \( P(x) \)**:
Since \( P(x) = f(x) + x \), we have:
\[ P(x) = \frac{1}{6}(x - 1)(x - 2)(x - 3) + x \]

7. **Calculate \( P(6) \)**:
Now we can find \( P(6) \):
\[ P(6) = \frac{1}{6}(6 - 1)(6 - 2)(6 - 3) + 6 \]
\[ = \frac{1}{6}(5)(4)(3) + 6 \]
\[ = \frac{60}{6} + 6 = 10 + 6 = 16 \]

### Final Answer:
Thus, \( P(6) = 16 \).