Let us consider the equation `cos^4x/a+sin^4x/b=1/(a+b),x in[0,pi/2],a,bgt0`
the value of `sin^8x/b^3+cos^8x/a^3` is

Options

A. `1/((a+b)^2)`
B. `1/((a+b)^3)`
C. `1/((a+b)^4)`
D. `1/(a^3+b^3)`

Explanation

To solve the equation

\[ \frac{\cos^4 x}{a} + \frac{\sin^4 x}{b} = \frac{1}{a+b} \]

for \(x \in [0, \frac{\pi}{2}]\) and given \(a, b > 0\), we need to find the value of

\[ \frac{\sin^8 x}{b^3} + \frac{\cos^8 x}{a^3}. \]

### Step 1: Rewrite the given equation

We start with the equation:

\[ \frac{\cos^4 x}{a} + \frac{\sin^4 x}{b} = \frac{1}{a+b}. \]

Multiply both sides by \(a + b\):

\[ (a + b) \left( \frac{\cos^4 x}{a} + \frac{\sin^4 x}{b} \right) = 1. \]

This simplifies to:

\[ \frac{(a + b) \cos^4 x}{a} + \frac{(a + b) \sin^4 x}{b} = 1. \]

### Step 2: Simplify the equation

This can be expressed as:

\[ \frac{b \cos^4 x + a \sin^4 x}{ab} = 1. \]

Thus, we have:

\[ b \cos^4 x + a \sin^4 x = ab. \]

### Step 3: Express \(\sin^2 x\) and \(\cos^2 x\)

Let \(s = \sin^2 x\) and \(c = \cos^2 x\). Then \(s + c = 1\). We can rewrite the equation as:

\[ b c^2 + a s^2 = ab. \]

Substituting \(s = 1 - c\):

\[ b c^2 + a (1 - c)^2 = ab. \]

Expanding this gives:

\[ b c^2 + a (1 - 2c + c^2) = ab. \]

Combining like terms results in:

\[ (b + a) c^2 - 2ac + a = ab. \]

### Step 4: Rearranging the equation

Rearranging gives us a quadratic in \(c\):

\[ (b + a) c^2 - 2ac + (a - ab) = 0. \]

### Step 5: Solve for \(s\) and \(c\)

Using the quadratic formula \(c = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}\):

\[ c = \frac{2a \pm \sqrt{(2a)^2 - 4(b + a)(a - ab)}}{2(b + a)}. \]

### Step 6: Find \(\sin^8 x\) and \(\cos^8 x\)

Now, we need to find:

\[ \frac{\sin^8 x}{b^3} + \frac{\cos^8 x}{a^3} = \frac{s^4}{b^3} + \frac{c^4}{a^3}. \]

Using \(s = \frac{b}{a+b}\) and \(c = \frac{a}{a+b}\):

\[ s^4 = \left(\frac{b}{a+b}\right)^4 \quad \text{and} \quad c^4 = \left(\frac{a}{a+b}\right)^4. \]

Thus,

\[ \frac{s^4}{b^3} + \frac{c^4}{a^3} = \frac{b^4}{(a+b)^4 b^3} + \frac{a^4}{(a+b)^4 a^3} = \frac{b}{(a+b)^4} + \frac{a}{(a+b)^4} = \frac{a+b}{(a+b)^4} = \frac{1}{(a+b)^3}. \]

### Final Answer

Thus, the value of

\[ \frac{\sin^8 x}{b^3} + \frac{\cos^8 x}{a^3} = \frac{1}{(a+b)^3}. \]