`sin^(6)A + cos^(6)A + 3sin^(2)A.cos^(2)A=`

Options

A. 0
B. 1
C. 2
D. 3

Answer

To solve the expression \( \sin^6 A + \cos^6 A + 3 \sin^2 A \cos^2 A \), we can use the identity for the sum of cubes. Here’s a step-by-step solution:

### Step 1: Recognize the sum of cubes
The expression \( \sin^6 A + \cos^6 A \) can be recognized as a sum of cubes. We can rewrite it using the identity:
\[ a^3 + b^3 = (a + b)(a^2 - ab + b^2) \]
where \( a = \sin^2 A \) and \( b = \cos^2 A \).

### Step 2: Apply the identity
Using the identity, we have:
\[ \sin^6 A + \cos^6 A = (\sin^2 A + \cos^2 A)((\sin^2 A)^2 - \sin^2 A \cos^2 A + (\cos^2 A)^2) \]
Since \( \sin^2 A + \cos^2 A = 1 \), we can simplify:
\[ \sin^6 A + \cos^6 A = 1 \cdot ((\sin^2 A)^2 - \sin^2 A \cos^2 A + (\cos^2 A)^2) \]

### Step 3: Simplify the remaining expression
Now, we need to simplify \( (\sin^2 A)^2 + (\cos^2 A)^2 - \sin^2 A \cos^2 A \):
\[ (\sin^2 A)^2 + (\cos^2 A)^2 = \sin^4 A + \cos^4 A \]
Using the identity \( \sin^4 A + \cos^4 A = (\sin^2 A + \cos^2 A)^2 - 2 \sin^2 A \cos^2 A \):
\[ \sin^4 A + \cos^4 A = 1^2 - 2 \sin^2 A \cos^2 A = 1 - 2 \sin^2 A \cos^2 A \]
Thus, we have:
\[ \sin^6 A + \cos^6 A = (1 - 2 \sin^2 A \cos^2 A) - \sin^2 A \cos^2 A \]
This simplifies to:
\[ \sin^6 A + \cos^6 A = 1 - 3 \sin^2 A \cos^2 A \]

### Step 4: Combine with the original expression
Now, we can combine this with the \( 3 \sin^2 A \cos^2 A \) from the original expression:
\[ \sin^6 A + \cos^6 A + 3 \sin^2 A \cos^2 A = (1 - 3 \sin^2 A \cos^2 A) + 3 \sin^2 A \cos^2 A \]
This simplifies to:
\[ \sin^6 A + \cos^6 A + 3 \sin^2 A \cos^2 A = 1 \]

### Final Answer
Thus, the final result is:
\[ \sin^6 A + \cos^6 A + 3 \sin^2 A \cos^2 A = 1 \]