If `sin^(6)theta+cos^(6) theta-1= lambda sin^(2) theta cos^(2) theta`, find the value of `lambda`.

Answer

To find the value of \( \lambda \) in the equation

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

we can follow these steps:

### Step 1: Rewrite the left-hand side
We can express \( \sin^6 \theta + \cos^6 \theta \) using the identity for the sum of cubes:

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

Using the identity \( a^3 + b^3 = (a + b)(a^2 - ab + b^2) \), where \( a = \sin^2 \theta \) and \( b = \cos^2 \theta \), we have:

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

### Step 2: Simplify using the Pythagorean identity
Since \( \sin^2 \theta + \cos^2 \theta = 1 \), we can simplify:

\[ \sin^6 \theta + \cos^6 \theta = 1 \cdot \left((\sin^2 \theta)^2 - \sin^2 \theta \cos^2 \theta + (\cos^2 \theta)^2\right). \]

### Step 3: Expand the expression
Now we need to expand \( (\sin^2 \theta)^2 + (\cos^2 \theta)^2 \):

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

Using the identity \( \sin^4 \theta + \cos^4 \theta = (\sin^2 \theta + \cos^2 \theta)^2 - 2\sin^2 \theta \cos^2 \theta = 1 - 2\sin^2 \theta \cos^2 \theta \):

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

### Step 4: Substitute back into the original equation
Now substituting this back into the original equation:

\[ (1 - 3\sin^2 \theta \cos^2 \theta) - 1 = \lambda \sin^2 \theta \cos^2 \theta. \]

This simplifies to:

\[ -3\sin^2 \theta \cos^2 \theta = \lambda \sin^2 \theta \cos^2 \theta. \]

### Step 5: Solve for \( \lambda \)
Assuming \( \sin^2 \theta \cos^2 \theta \neq 0 \), we can divide both sides by \( \sin^2 \theta \cos^2 \theta \):

\[ \lambda = -3. \]

Thus, the value of \( \lambda \) is

\[ \boxed{-3}. \]