If `alpha` satisfies `15 sin^(4)alpha + 10 cos^(4)alpha=6`, then find the value of `27 sin^(8)alpha + 8 cos^(8) alpha`

Explanation

To solve the problem, we start with the given equation:

\[ 15 \sin^4 \alpha + 10 \cos^4 \alpha = 6 \]

### Step 1: Rewrite the equation
We can express \( \sin^4 \alpha \) and \( \cos^4 \alpha \) in terms of \( \sin^2 \alpha \) and \( \cos^2 \alpha \).

Using the identity \( \sin^2 \alpha + \cos^2 \alpha = 1 \), we can rewrite \( \cos^4 \alpha \) as:

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

Substituting this into the original equation gives:

\[ 15 \sin^4 \alpha + 10 (1 - 2\sin^2 \alpha + \sin^4 \alpha) = 6 \]

### Step 2: Simplify the equation
Expanding the equation:

\[ 15 \sin^4 \alpha + 10 - 20 \sin^2 \alpha + 10 \sin^4 \alpha = 6 \]

Combining like terms:

\[ (15 + 10) \sin^4 \alpha - 20 \sin^2 \alpha + 10 - 6 = 0 \]

This simplifies to:

\[ 25 \sin^4 \alpha - 20 \sin^2 \alpha + 4 = 0 \]

### Step 3: Substitute \( x = \sin^2 \alpha \)
Let \( x = \sin^2 \alpha \). Then the equation becomes:

\[ 25x^2 - 20x + 4 = 0 \]

### Step 4: Solve the quadratic equation
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):

Here, \( a = 25 \), \( b = -20 \), and \( c = 4 \).

Calculating the discriminant:

\[ b^2 - 4ac = (-20)^2 - 4 \cdot 25 \cdot 4 = 400 - 400 = 0 \]

Since the discriminant is zero, there is one solution:

\[ x = \frac{20}{2 \cdot 25} = \frac{20}{50} = \frac{2}{5} \]

### Step 5: Find \( \sin^2 \alpha \) and \( \cos^2 \alpha \)
Thus, we have:

\[ \sin^2 \alpha = \frac{2}{5} \]

Using \( \sin^2 \alpha + \cos^2 \alpha = 1 \):

\[ \cos^2 \alpha = 1 - \sin^2 \alpha = 1 - \frac{2}{5} = \frac{3}{5} \]

### Step 6: Calculate \( 27 \sin^8 \alpha + 8 \cos^8 \alpha \)
Now we need to calculate:

\[ 27 \sin^8 \alpha + 8 \cos^8 \alpha \]

Calculating \( \sin^8 \alpha \) and \( \cos^8 \alpha \):

\[ \sin^8 \alpha = \left(\sin^2 \alpha\right)^4 = \left(\frac{2}{5}\right)^4 = \frac{16}{625} \]

\[ \cos^8 \alpha = \left(\cos^2 \alpha\right)^4 = \left(\frac{3}{5}\right)^4 = \frac{81}{625} \]

Now substituting back into the expression:

\[ 27 \sin^8 \alpha + 8 \cos^8 \alpha = 27 \cdot \frac{16}{625} + 8 \cdot \frac{81}{625} \]

Calculating each term:

\[ = \frac{432}{625} + \frac{648}{625} = \frac{432 + 648}{625} = \frac{1080}{625} \]

### Final Result
Thus, the value of \( 27 \sin^8 \alpha + 8 \cos^8 \alpha \) is:

\[ \frac{1080}{625} \]