The value of `2(sin^6 theta+ cos^6 theta )-3 (sin^4 theta + cos^4 theta)` is ……..

Answer

To solve the expression \(2(\sin^6 \theta + \cos^6 \theta) - 3(\sin^4 \theta + \cos^4 \theta)\), we can break it down step by step.

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

### Step 2: Express \(\sin^6 \theta + \cos^6 \theta\)
We can use 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: Express \(\sin^4 \theta + \cos^4 \theta\)
Using the identity:
\[ \sin^4 \theta + \cos^4 \theta = (\sin^2 \theta + \cos^2 \theta)^2 - 2\sin^2 \theta \cos^2 \theta \]
This simplifies to:
\[ \sin^4 \theta + \cos^4 \theta = 1 - 2\sin^2 \theta \cos^2 \theta \]

### Step 4: Substitute back into the original expression
Now substituting these results back into the original expression:
\[ 2(\sin^6 \theta + \cos^6 \theta) - 3(\sin^4 \theta + \cos^4 \theta) = 2(\sin^4 \theta - \sin^2 \theta \cos^2 \theta + \cos^4 \theta) - 3(1 - 2\sin^2 \theta \cos^2 \theta) \]
This simplifies to:
\[ = 2(\sin^4 \theta + \cos^4 \theta) - 2\sin^2 \theta \cos^2 \theta - 3 + 6\sin^2 \theta \cos^2 \theta \]
Combining like terms gives:
\[ = 2(\sin^4 \theta + \cos^4 \theta) + 4\sin^2 \theta \cos^2 \theta - 3 \]

### Step 5: Substitute \(\sin^4 \theta + \cos^4 \theta\)
Substituting \(\sin^4 \theta + \cos^4 \theta = 1 - 2\sin^2 \theta \cos^2 \theta\):
\[ = 2(1 - 2\sin^2 \theta \cos^2 \theta) + 4\sin^2 \theta \cos^2 \theta - 3 \]
This simplifies to:
\[ = 2 - 4\sin^2 \theta \cos^2 \theta + 4\sin^2 \theta \cos^2 \theta - 3 \]
Thus:
\[ = 2 - 3 = -1 \]

### Final Answer
The value of the expression is:
\[ \boxed{-1} \]