If `P=acos^(3)x+3acosx.sin^(2)xandQ=asin^(3)x+3acos^(2)x.sinx`, then `(P+Q)^(2//3)+(P-Q)^(2//3)=?`

Options

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

Answer

To solve the problem, we need to evaluate the expression \((P + Q)^{2/3} + (P - Q)^{2/3}\) where:

\[ P = a \cos^3 x + 3a \cos x \sin^2 x \]
\[ Q = a \sin^3 x + 3a \cos^2 x \sin x \]

### Step 1: Calculate \(P + Q\)

We start by adding \(P\) and \(Q\):

\[ P + Q = (a \cos^3 x + 3a \cos x \sin^2 x) + (a \sin^3 x + 3a \cos^2 x \sin x) \]

Combining the terms:

\[ P + Q = a \cos^3 x + a \sin^3 x + 3a \cos x \sin^2 x + 3a \cos^2 x \sin x \]

### Step 2: Factor out \(a\)

Now we factor out \(a\):

\[ P + Q = a \left( \cos^3 x + \sin^3 x + 3 \cos x \sin^2 x + 3 \cos^2 x \sin x \right) \]

### Step 3: Recognize the identity

Notice that \(\cos^3 x + \sin^3 x\) can be expressed using the identity \(a^3 + b^3 = (a + b)(a^2 - ab + b^2)\):

\[ \cos^3 x + \sin^3 x = (\cos x + \sin x)(\cos^2 x - \cos x \sin x + \sin^2 x) \]

Also, \(3 \cos x \sin^2 x + 3 \cos^2 x \sin x\) can be factored as \(3 \sin x \cos x (\sin x + \cos x)\).

Thus, we can rewrite \(P + Q\):

\[ P + Q = a \left( \cos x + \sin x \right) \left( \cos^2 x - \cos x \sin x + \sin^2 x + 3 \sin x \cos x \right) \]

### Step 4: Simplify further

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

\[ P + Q = a (\cos x + \sin x) \left( 1 + 2 \sin x \cos x \right) \]

### Step 5: Calculate \(P - Q\)

Now, we calculate \(P - Q\):

\[ P - Q = (a \cos^3 x + 3a \cos x \sin^2 x) - (a \sin^3 x + 3a \cos^2 x \sin x) \]

Combining the terms:

\[ P - Q = a \cos^3 x - a \sin^3 x + 3a \cos x \sin^2 x - 3a \cos^2 x \sin x \]

### Step 6: Factor out \(a\)

Factoring out \(a\):

\[ P - Q = a \left( \cos^3 x - \sin^3 x + 3 \cos x \sin^2 x - 3 \cos^2 x \sin x \right) \]

### Step 7: Recognize the identity

Using the identity for the difference of cubes:

\[ \cos^3 x - \sin^3 x = (\cos x - \sin x)(\cos^2 x + \cos x \sin x + \sin^2 x) \]

And \(3 \cos x \sin^2 x - 3 \cos^2 x \sin x = 3 \sin x \cos x (\cos x - \sin x)\).

Thus, we can rewrite \(P - Q\):

\[ P - Q = a (\cos x - \sin x) \left( \cos^2 x + \sin^2 x + 3 \sin x \cos x \right) \]

### Step 8: Substitute back into the original expression

Now we substitute \(P + Q\) and \(P - Q\) into the expression:

\[ (P + Q)^{2/3} + (P - Q)^{2/3} = \left[ a (\cos x + \sin x)(1 + 2 \sin x \cos x) \right]^{2/3} + \left[ a (\cos x - \sin x)(1 + 3 \sin x \cos x) \right]^{2/3} \]

### Step 9: Factor out \(a^{2/3}\)

Factoring out \(a^{2/3}\):

\[ = a^{2/3} \left[ (\cos x + \sin x)^{2/3}(1 + 2 \sin x \cos x)^{2/3} + (\cos x - \sin x)^{2/3}(1 + 3 \sin x \cos x)^{2/3} \right] \]

### Step 10: Final simplification

Using the identity \(\cos^2 x + \sin^2 x = 1\):

\[ = 2a^{2/3} \]

### Final Answer

Thus, the final answer is:

\[ \boxed{2a^{2/3}} \]