If x^2+y^2= 1,then (4x^3-3x)^2+(3y-4y^3)^2

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Solution

Given : x² + y² = 1 -------(1)

then,
multiply equ. (1) by 4.

4x² + 4y² = 4
=> 4x² = 4 - 4y² and
4y² = 4 - 4x²

* indicate multiplication.

(4x³ - 3x)² + (3y - 4y³)² = [ x * (4x² - 3) ]² + [ y * (3 - 4y²) ]²

=> x² * (4 - 4y² - 3)² + y² * (3 - 4 + 4x²)²
=> x² * (1 - 4y²)² + y² * (4x² - 1)²
=> x² * (1 + 16y^4 - 8y² ) + y² * (16x^4 + 1 - 8x²)
=> x² + 16x²y^4 - 8x²y² + 16x^4 y² + y² - 8x²y²
=> (x² + y²) + 16x²y² ( x² + y²) - 16x²y²
=> 1 + 16x²y² - 16x²y² . . . {As x² + y² = 1 }

=> 1

answer = 1

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