The value of the expression `cos^6theta+sin^6theta+3sin^2thetacos^2theta=`

Options

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

Answer

To solve the expression \( \cos^6 \theta + \sin^6 \theta + 3 \sin^2 \theta \cos^2 \theta \), we can use the identity related to the sum of cubes. Let's break it down step by step.

### Step 1: Identify the structure
The expression \( \cos^6 \theta + \sin^6 \theta + 3 \sin^2 \theta \cos^2 \theta \) resembles the identity for the sum of cubes, which is:
\[ a^3 + b^3 + 3ab(a + b) = (a + b)^3 \]
where \( a = \sin^2 \theta \) and \( b = \cos^2 \theta \).

### Step 2: Assign values to \( a \) and \( b \)
Let:
\[ a = \sin^2 \theta \quad \text{and} \quad b = \cos^2 \theta \]

### Step 3: Rewrite the expression
Now we can rewrite the expression:
\[ \cos^6 \theta + \sin^6 \theta + 3 \sin^2 \theta \cos^2 \theta = a^3 + b^3 + 3ab \]

### Step 4: Apply the identity
Using the identity:
\[ a^3 + b^3 + 3ab = (a + b)^3 \]
we can substitute \( a + b \):
\[ \sin^2 \theta + \cos^2 \theta = 1 \]
Thus, we have:
\[ a^3 + b^3 + 3ab = (1)^3 = 1 \]

### Step 5: Conclusion
Therefore, the value of the expression \( \cos^6 \theta + \sin^6 \theta + 3 \sin^2 \theta \cos^2 \theta \) is:
\[ \boxed{1} \]

---