`sin^(6)A+cos^(6)A` is equal to :

Options

A. `1-3sin^(2)A cos^(2)A`
B. `1-3sinAcosA`
C. `1+3sin^(2)Acos^(2)A`
D. 1

Answer

To solve the expression \( \sin^6 A + \cos^6 A \), we can use the identities and formulas related to cubes and squares. Here’s a step-by-step breakdown of the solution:

### Step 1: Recognize the Form
We can rewrite \( \sin^6 A + \cos^6 A \) as \( ( \sin^2 A )^3 + ( \cos^2 A )^3 \). Let \( x = \sin^2 A \) and \( y = \cos^2 A \). Thus, we have:
\[ x^3 + y^3 \]

### Step 2: Apply the Sum of Cubes Formula
The sum of cubes can be factored using the formula:
\[ a^3 + b^3 = (a + b)(a^2 - ab + b^2) \]
Applying this to our expression:
\[ x^3 + y^3 = (x + y)(x^2 - xy + y^2) \]

### Step 3: Substitute Values
From the Pythagorean identity, we know:
\[ x + y = \sin^2 A + \cos^2 A = 1 \]
Thus, we can substitute this into our equation:
\[ x^3 + y^3 = 1 \cdot (x^2 - xy + y^2) = x^2 - xy + y^2 \]

### Step 4: Simplify \( x^2 - xy + y^2 \)
Now we need to simplify \( x^2 - xy + y^2 \):
\[ x^2 + y^2 = (\sin^2 A)^2 + (\cos^2 A)^2 \]
Using the identity \( x^2 + y^2 = (x + y)^2 - 2xy \):
\[ x^2 + y^2 = 1^2 - 2xy = 1 - 2xy \]
Thus, we have:
\[ x^2 - xy + y^2 = (1 - 2xy) - xy = 1 - 3xy \]

### Step 5: Final Expression
Now, substituting back, we get:
\[ \sin^6 A + \cos^6 A = 1 - 3 \sin^2 A \cos^2 A \]

### Step 6: Conclusion
The final answer is:
\[ \sin^6 A + \cos^6 A = 1 - 3 \sin^2 A \cos^2 A \]