If `x^(2) + y^(2) =1` and `P = ( 3x -4x^(3))^(2) + ( 3y - 4y^(3) ) ^(2)`, then `P ="……………."`

Answer

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

1. \( x^2 + y^2 = 1 \)
2. \( P = (3x - 4x^3)^2 + (3y - 4y^3)^2 \)

### Step 1: Expand the expression for \( P \)

We need to expand \( P \):

\[ P = (3x - 4x^3)^2 + (3y - 4y^3)^2 \]

Using the formula \( (a - b)^2 = a^2 - 2ab + b^2 \):

\[ P = (3x)^2 - 2(3x)(4x^3) + (4x^3)^2 + (3y)^2 - 2(3y)(4y^3) + (4y^3)^2 \]

Calculating each term:

\[ P = 9x^2 - 24x^4 + 16x^6 + 9y^2 - 24y^4 + 16y^6 \]

### Step 2: Combine like terms

Now, we can combine the terms:

\[ P = (9x^2 + 9y^2) + (16x^6 + 16y^6) - (24x^4 + 24y^4) \]

Using the fact that \( x^2 + y^2 = 1 \):

\[ P = 9(1) + 16(x^6 + y^6) - 24(x^4 + y^4) \]

### Step 3: Express \( x^6 + y^6 \) and \( x^4 + y^4 \)

To express \( x^6 + y^6 \) and \( x^4 + y^4 \) in terms of \( x^2 + y^2 \):

Using the identity:

\[ x^6 + y^6 = (x^2 + y^2)(x^4 + y^4) - x^2y^2(x^2 + y^2) \]

Let \( s = x^2 + y^2 = 1 \) and \( p = x^2y^2 \):

\[ x^4 + y^4 = (x^2 + y^2)^2 - 2x^2y^2 = s^2 - 2p = 1 - 2p \]

Substituting back:

\[ x^6 + y^6 = s(1 - 2p) - ps = 1 - 3p \]

### Step 4: Substitute back into \( P \)

Now substituting \( x^6 + y^6 \) and \( x^4 + y^4 \) back into \( P \):

\[ P = 9 + 16(1 - 3p) - 24(1 - 2p) \]

Expanding this:

\[ P = 9 + 16 - 48p - 24 + 48p \]

### Step 5: Simplify \( P \)

Now simplifying:

\[ P = 9 + 16 - 24 = 1 \]

### Final Answer

Thus, the value of \( P \) is:

\[ \boxed{1} \]