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

Options

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

Answer

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

### Step 1: Recognize the identity
We can use the identity for the sum of cubes:
\[ a^3 + b^3 = (a + b)(a^2 - ab + b^2) \]
In our case, let \( a = \sin^2 A \) and \( b = \cos^2 A \). Thus, we can rewrite \( \sin^6 A + \cos^6 A \) as:
\[ (\sin^2 A)^3 + (\cos^2 A)^3 \]

### Step 2: Apply the sum of cubes formula
Using the sum of cubes identity:
\[ \sin^6 A + \cos^6 A = (\sin^2 A + \cos^2 A)((\sin^2 A)^2 - \sin^2 A \cos^2 A + (\cos^2 A)^2) \]

### Step 3: Simplify using the Pythagorean identity
We know from the Pythagorean identity that:
\[ \sin^2 A + \cos^2 A = 1 \]
Thus, substituting this into our equation gives:
\[ \sin^6 A + \cos^6 A = 1 \cdot ((\sin^2 A)^2 - \sin^2 A \cos^2 A + (\cos^2 A)^2) \]

### Step 4: Simplify the remaining expression
Now we need to simplify \( (\sin^2 A)^2 + (\cos^2 A)^2 - \sin^2 A \cos^2 A \):
\[ (\sin^2 A)^2 + (\cos^2 A)^2 = \sin^4 A + \cos^4 A \]
We can also use the identity:
\[ \sin^4 A + \cos^4 A = (\sin^2 A + \cos^2 A)^2 - 2\sin^2 A \cos^2 A \]
Substituting \( \sin^2 A + \cos^2 A = 1 \):
\[ \sin^4 A + \cos^4 A = 1 - 2\sin^2 A \cos^2 A \]

### Step 5: Substitute back into the equation
Now substituting this back into our expression:
\[ \sin^6 A + \cos^6 A = 1 - 2\sin^2 A \cos^2 A - \sin^2 A \cos^2 A \]
This simplifies to:
\[ \sin^6 A + \cos^6 A = 1 - 3\sin^2 A \cos^2 A \]

### Final Result
Thus, the final expression for \( \sin^6 A + \cos^6 A \) is:
\[ \sin^6 A + \cos^6 A = 1 - 3\sin^2 A \cos^2 A \]

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