Step 1: Understanding the Concept:
This problem requires simplifying a trigonometric expression using algebraic identities and the fundamental Pythagorean identity $\sin^2\theta + \cos^2\theta = 1$.
Step 2: Key Formula or Approach:
We can view $\sin^6\theta + \cos^6\theta$ as a sum of cubes, $(\sin^2\theta)^3 + (\cos^2\theta)^3$. We will use the algebraic identity:
\[ a^3 + b^3 = (a+b)(a^2 - ab + b^2) \]
or, more conveniently:
\[ a^3 + b^3 = (a+b)^3 - 3ab(a+b) \]
Let $a = \sin^2\theta$ and $b = \cos^2\theta$.
Step 3: Detailed Explanation:
Let $a = \sin^2\theta$ and $b = \cos^2\theta$. We know that $a+b = \sin^2\theta + \cos^2\theta = 1$.
The expression is $a^3 + b^3 + 3ab$.
Using the second identity from Step 2:
\[ a^3 + b^3 = (a+b)^3 - 3ab(a+b) \]
Substitute the values of $a$ and $b$:
\[ \sin^6\theta + \cos^6\theta = (\sin^2\theta + \cos^2\theta)^3 - 3(\sin^2\theta)(\cos^2\theta)(\sin^2\theta + \cos^2\theta) \]
Since $\sin^2\theta + \cos^2\theta = 1$:
\[ \sin^6\theta + \cos^6\theta = (1)^3 - 3\sin^2\theta\cos^2\theta(1) \]
\[ \sin^6\theta + \cos^6\theta = 1 - 3\sin^2\theta\cos^2\theta \]
Now, substitute this result back into the original expression given in the question:
\[ (\sin^6\theta + \cos^6\theta) + 3\sin^2\theta \cos^2\theta = (1 - 3\sin^2\theta\cos^2\theta) + 3\sin^2\theta \cos^2\theta \]
\[ = 1 \]
Step 4: Final Answer:
The value of the expression simplifies to 1. Therefore, option (B) is the correct answer.