Question

X^2+y^2=1then (4x^3-3x)^2+(3y-4y^3)^2​

Answer 1

Answer:

hey mate up there is your answer

Answer 2

Answer:

(4x³- 3x)² + (3y - 4y³)² = 1

Step-by-step explanation:

Given: x² + y² = 1

We need to find the value of (4x³- 3x)² + (3y - 4y³)²

On expanding,

= 16x⁶ + 9x²- 24x⁴ + 9y² + 16y⁶ - 24y⁴

= 16 (x⁶ + y⁶) + 9(x² + y²) - 24(x⁴ + y⁴)                 --(i)

As (x² + y²)³ = x⁶ + y⁶ + 3x²y²(x² + y²)

and (x² + y²)² = x⁴ + y⁴ + 2x²y²

Substituting these values in equation (i)

= 16 [(x² + y²)³- 3x²y²(x² + y²) ]+ 9(x² + y²) - 24[(x² + y²)² - 2x²y²]  

As  x² + y² = 1

=>  16 [(1)³- 3x²y²(1) ]+ 9(1) - 24[(1)² - 2x²y²]

=> 16 [1- 3x²y² ]+ 9 - 24[1 - 2x²y²]

=> 16 - 48x²y² + 9 - 24 + 48x²y²

=> 25 - 24

= 1

Therefore, (4x³- 3x)² + (3y - 4y³)² = 1