What is the value of `sin^6 theta + cos^6 theta + 3sin^2 theta cos^2 theta` ?

Answer

To find the value of the expression \( \sin^6 \theta + \cos^6 \theta + 3 \sin^2 \theta \cos^2 \theta \), we can follow these steps:

### Step 1: Use the identity for \( \sin^2 \theta + \cos^2 \theta \)
We know that:
\[ \sin^2 \theta + \cos^2 \theta = 1 \]

### Step 2: Apply the identity for the cube of a binomial
We can express \( \sin^6 \theta + \cos^6 \theta \) using the identity for the sum of cubes:
\[ a^3 + b^3 = (a + b)(a^2 - ab + b^2) \]
Let \( a = \sin^2 \theta \) and \( b = \cos^2 \theta \). Then,
\[ \sin^6 \theta + \cos^6 \theta = (\sin^2 \theta + \cos^2 \theta)(\sin^4 \theta - \sin^2 \theta \cos^2 \theta + \cos^4 \theta) \]
Since \( \sin^2 \theta + \cos^2 \theta = 1 \), we have:
\[ \sin^6 \theta + \cos^6 \theta = \sin^4 \theta - \sin^2 \theta \cos^2 \theta + \cos^4 \theta \]

### Step 3: Simplify \( \sin^4 \theta + \cos^4 \theta \)
We can further simplify \( \sin^4 \theta + \cos^4 \theta \) using the identity:
\[ \sin^4 \theta + \cos^4 \theta = (\sin^2 \theta + \cos^2 \theta)^2 - 2\sin^2 \theta \cos^2 \theta \]
Substituting \( \sin^2 \theta + \cos^2 \theta = 1 \):
\[ \sin^4 \theta + \cos^4 \theta = 1^2 - 2\sin^2 \theta \cos^2 \theta = 1 - 2\sin^2 \theta \cos^2 \theta \]

### Step 4: Substitute back into the expression
Now we substitute back into our earlier expression:
\[ \sin^6 \theta + \cos^6 \theta = (1 - 2\sin^2 \theta \cos^2 \theta) - \sin^2 \theta \cos^2 \theta \]
This simplifies to:
\[ \sin^6 \theta + \cos^6 \theta = 1 - 3\sin^2 \theta \cos^2 \theta \]

### Step 5: Combine with the original expression
Now we substitute this back into the original expression:
\[ \sin^6 \theta + \cos^6 \theta + 3\sin^2 \theta \cos^2 \theta = (1 - 3\sin^2 \theta \cos^2 \theta) + 3\sin^2 \theta \cos^2 \theta \]
This simplifies to:
\[ 1 - 3\sin^2 \theta \cos^2 \theta + 3\sin^2 \theta \cos^2 \theta = 1 \]

### Final Answer
Thus, the value of \( \sin^6 \theta + \cos^6 \theta + 3\sin^2 \theta \cos^2 \theta \) is:
\[ \boxed{1} \]