The value of `[(sqrt3 + 2sinP)/ (1- 2Cos P)]^3 + [(1 + 2Cos P )/(sqrt3-2sinP)]^3` is:

Options

A. `Sin PCosP`
B. 1
C. `2Sin PCosP`
D. 0

Answer

To solve the problem, we need to evaluate the expression:

\[ \left(\frac{\sqrt{3} + 2\sin P}{1 - 2\cos P}\right)^3 + \left(\frac{1 + 2\cos P}{\sqrt{3} - 2\sin P}\right)^3 \]

### Step 1: Substitute a specific value for \( P \)

Let's substitute \( P = 30^\circ \) (or \( \frac{\pi}{6} \) radians) since it is a common angle that simplifies calculations.

### Step 2: Calculate \( \sin P \) and \( \cos P \)

For \( P = 30^\circ \):
- \( \sin 30^\circ = \frac{1}{2} \)
- \( \cos 30^\circ = \frac{\sqrt{3}}{2} \)

### Step 3: Substitute the values into the expression

Now, substitute these values into the expression:

1. For the first term:
\[ \frac{\sqrt{3} + 2\sin 30^\circ}{1 - 2\cos 30^\circ} = \frac{\sqrt{3} + 2 \cdot \frac{1}{2}}{1 - 2 \cdot \frac{\sqrt{3}}{2}} = \frac{\sqrt{3} + 1}{1 - \sqrt{3}} \]

2. For the second term:
\[ \frac{1 + 2\cos 30^\circ}{\sqrt{3} - 2\sin 30^\circ} = \frac{1 + 2 \cdot \frac{\sqrt{3}}{2}}{\sqrt{3} - 2 \cdot \frac{1}{2}} = \frac{1 + \sqrt{3}}{\sqrt{3} - 1} \]

### Step 4: Rewrite the expression

Now we can rewrite the entire expression using the results from above:

\[ \left(\frac{\sqrt{3} + 1}{1 - \sqrt{3}}\right)^3 + \left(\frac{1 + \sqrt{3}}{\sqrt{3} - 1}\right)^3 \]

### Step 5: Recognize a pattern

Notice that:
\[ \frac{\sqrt{3} + 1}{1 - \sqrt{3}} = -\frac{1 + \sqrt{3}}{\sqrt{3} - 1} \]
This means that the two fractions are negatives of each other.

### Step 6: Apply the identity for cubes

Using the identity \( a^3 + b^3 = (a + b)(a^2 - ab + b^2) \):
Let \( a = \frac{\sqrt{3} + 1}{1 - \sqrt{3}} \) and \( b = \frac{1 + \sqrt{3}}{\sqrt{3} - 1} \).

Since \( a + b = 0 \), we have:
\[ a^3 + b^3 = 0 \]

### Conclusion

Thus, the value of the original expression is:

\[ \left(\frac{\sqrt{3} + 2\sin P}{1 - 2\cos P}\right)^3 + \left(\frac{1 + 2\cos P}{\sqrt{3} - 2\sin P}\right)^3 = 0 \]

So the answer is:

**0**