If `sin^(2)alpha,cos^(2)alphaand-cosec^(2)alpha` are the zeros of `P(x)=x^(3)+x^(2)+ax+b(a,binR)`. Then P(2) equals ______.

Explanation

To solve the problem, we need to find the value of \( P(2) \) for the polynomial \( P(x) = x^3 + x^2 + ax + b \), given that the zeros of the polynomial are \( \sin^2 \alpha \), \( \cos^2 \alpha \), and \( -\csc^2 \alpha \).

### Step 1: Identify the roots and their relationships
The roots of the polynomial are:
- \( r_1 = \sin^2 \alpha \)
- \( r_2 = \cos^2 \alpha \)
- \( r_3 = -\csc^2 \alpha \)

From the Pythagorean identity, we know:
\[ \sin^2 \alpha + \cos^2 \alpha = 1 \]

### Step 2: Calculate the sum of the roots
Using Vieta's formulas, the sum of the roots \( r_1 + r_2 + r_3 \) is equal to \( -\frac{\text{coefficient of } x^2}{\text{coefficient of } x^3} \):
\[ r_1 + r_2 + r_3 = -1 \]
Substituting the known values:
\[ \sin^2 \alpha + \cos^2 \alpha - \csc^2 \alpha = -1 \]
Since \( \sin^2 \alpha + \cos^2 \alpha = 1 \), we can substitute:
\[ 1 - \csc^2 \alpha = -1 \]
This simplifies to:
\[ -\csc^2 \alpha = -2 \implies \csc^2 \alpha = 2 \]
Thus, we have:
\[ \sin^2 \alpha = \frac{1}{2} \quad \text{and} \quad \cos^2 \alpha = \frac{1}{2} \]

### Step 3: Substitute the roots into the polynomial
Now we can express the polynomial \( P(x) \) in terms of its roots:
\[ P(x) = (x - \sin^2 \alpha)(x - \cos^2 \alpha)(x + \csc^2 \alpha) \]
Substituting the values of the roots:
\[ P(x) = \left(x - \frac{1}{2}\right)\left(x - \frac{1}{2}\right)\left(x + 2\right) \]
This simplifies to:
\[ P(x) = \left(x - \frac{1}{2}\right)^2 (x + 2) \]

### Step 4: Expand the polynomial
First, expand \( \left(x - \frac{1}{2}\right)^2 \):
\[ \left(x - \frac{1}{2}\right)^2 = x^2 - x + \frac{1}{4} \]
Now multiply by \( (x + 2) \):
\[ P(x) = (x^2 - x + \frac{1}{4})(x + 2) \]
Expanding this:
\[ P(x) = x^3 + 2x^2 - x^2 - 2x + \frac{1}{4}x + \frac{1}{2} \]
Combine like terms:
\[ P(x) = x^3 + x^2 - \frac{7}{4}x + \frac{1}{2} \]

### Step 5: Evaluate \( P(2) \)
Now we substitute \( x = 2 \):
\[ P(2) = 2^3 + 2^2 - \frac{7}{4}(2) + \frac{1}{2} \]
Calculating each term:
\[ = 8 + 4 - \frac{14}{4} + \frac{2}{4} \]
\[ = 8 + 4 - \frac{12}{4} \]
\[ = 8 + 4 - 3 = 9 \]

Thus, the final answer is:
\[ \boxed{9} \]