The value of the expression
`3(sin theta- cos theta)^4 + 6(sin theta + cos theta)^2 + 4(sin^6 theta+ cos^6 theta)`

Options

A. 1
B. `-1`
C. 13
D. 0

Answer

To find the value of the expression

\[ 3(\sin \theta - \cos \theta)^4 + 6(\sin \theta + \cos \theta)^2 + 4(\sin^6 \theta + \cos^6 \theta) \]

we will simplify each term step by step.

### Step 1: Expand \((\sin \theta - \cos \theta)^4\)

Using the binomial theorem, we can expand \((a - b)^4\):

\[ (\sin \theta - \cos \theta)^4 = \sin^4 \theta - 4\sin^3 \theta \cos \theta + 6\sin^2 \theta \cos^2 \theta - 4\sin \theta \cos^3 \theta + \cos^4 \theta \]

### Step 2: Expand \((\sin \theta + \cos \theta)^2\)

\[ (\sin \theta + \cos \theta)^2 = \sin^2 \theta + 2\sin \theta \cos \theta + \cos^2 \theta = 1 + 2\sin \theta \cos \theta \]

### Step 3: Calculate \(\sin^6 \theta + \cos^6 \theta\)

Using the identity for the sum of cubes:

\[ \sin^6 \theta + \cos^6 \theta = (\sin^2 \theta + \cos^2 \theta)(\sin^4 \theta - \sin^2 \theta \cos^2 \theta + \cos^4 \theta) = 1(\sin^4 \theta - \sin^2 \theta \cos^2 \theta + \cos^4 \theta) \]

We can rewrite \(\sin^4 \theta + \cos^4 \theta\) as:

\[ \sin^4 \theta + \cos^4 \theta = (\sin^2 \theta + \cos^2 \theta)^2 - 2\sin^2 \theta \cos^2 \theta = 1 - 2\sin^2 \theta \cos^2 \theta \]

Thus,

\[ \sin^6 \theta + \cos^6 \theta = 1 - 3\sin^2 \theta \cos^2 \theta \]

### Step 4: Substitute back into the expression

Now substituting everything back into the original expression:

\[ 3(\sin^4 \theta - 4\sin^3 \theta \cos \theta + 6\sin^2 \theta \cos^2 \theta - 4\sin \theta \cos^3 \theta + \cos^4 \theta) + 6(1 + 2\sin \theta \cos \theta) + 4(1 - 3\sin^2 \theta \cos^2 \theta) \]

### Step 5: Combine like terms

Combining all the terms, we simplify:

1. The constant terms: \(3 + 6 + 4 = 13\)
2. The \(\sin^2 \theta \cos^2 \theta\) terms: \(18\sin^2 \theta \cos^2 \theta - 12\sin^2 \theta \cos^2 \theta = 6\sin^2 \theta \cos^2 \theta\)
3. The terms involving \(\sin^3 \theta \cos \theta\) and \(\sin \theta \cos^3 \theta\) will cancel out or combine to zero.

### Final Result

Thus, the expression simplifies to:

\[ 13 + 6\sin^2 \theta \cos^2 \theta \]

Since \(\sin^2 \theta \cos^2 \theta\) can take values from 0 to 1, the minimum value of the expression is 13 when \(\sin^2 \theta \cos^2 \theta = 0\).

Therefore, the final value of the expression is:

\[ \boxed{13} \]