The maximum value of `y=1/(sin^6x+cos^6x)` is ______

Explanation

To find the maximum value of \( y = \frac{1}{\sin^6 x + \cos^6 x} \), we can follow these steps:

### Step 1: Rewrite the expression
We start with the expression:
\[ y = \frac{1}{\sin^6 x + \cos^6 x} \]
We can use the identity for \( a^3 + b^3 \):
\[ a^3 + b^3 = (a + b)(a^2 - ab + b^2) \]
Let \( a = \sin^2 x \) and \( b = \cos^2 x \). Then,
\[ \sin^6 x + \cos^6 x = (\sin^2 x)^3 + (\cos^2 x)^3 = (\sin^2 x + \cos^2 x)((\sin^2 x)^2 - \sin^2 x \cos^2 x + (\cos^2 x)^2) \]

### Step 2: Simplify using the Pythagorean identity
Since \( \sin^2 x + \cos^2 x = 1 \), we have:
\[ \sin^6 x + \cos^6 x = 1 \cdot ((\sin^2 x)^2 - \sin^2 x \cos^2 x + (\cos^2 x)^2) \]
Now, we need to simplify \( (\sin^2 x)^2 + (\cos^2 x)^2 \):
\[ (\sin^2 x)^2 + (\cos^2 x)^2 = \sin^4 x + \cos^4 x = (\sin^2 x + \cos^2 x)^2 - 2\sin^2 x \cos^2 x = 1 - 2\sin^2 x \cos^2 x \]
Thus,
\[ \sin^6 x + \cos^6 x = 1 - 3\sin^2 x \cos^2 x \]

### Step 3: Substitute back into the expression for \( y \)
Now we can rewrite \( y \):
\[ y = \frac{1}{1 - 3\sin^2 x \cos^2 x} \]

### Step 4: Use the double angle identity
Using the double angle identity, \( \sin 2x = 2\sin x \cos x \), we have:
\[ \sin^2 x \cos^2 x = \frac{1}{4} \sin^2 2x \]
Substituting this into our expression for \( y \):
\[ y = \frac{1}{1 - \frac{3}{4} \sin^2 2x} = \frac{4}{4 - 3\sin^2 2x} \]

### Step 5: Find the maximum value of \( y \)
To maximize \( y \), we need to minimize the denominator \( 4 - 3\sin^2 2x \). The minimum value of \( \sin^2 2x \) is 0, and the maximum value is 1. Thus, the minimum value of the denominator occurs when \( \sin^2 2x = 1 \):
\[ 4 - 3 \cdot 1 = 1 \]
Therefore, the maximum value of \( y \) is:
\[ y_{\text{max}} = \frac{4}{1} = 4 \]

### Final Answer
The maximum value of \( y = \frac{1}{\sin^6 x + \cos^6 x} \) is \( \boxed{4} \).

---