If x = ( √ 3 + 1 ) / ( √ 3 - 1 ) and y = ( √ 3 - 1) / ( √ 3 + 1 ) , then the value of x2/y + y2/x is?

Given, x = ( √3 + 1) / ( √3 - 1 )
y = ( √3 - 1 ) / ( √3 + 1 )
Then , x + y = { ( √3 + 1) / ( √3 - 1 ) } + { ( √3 - 1 ) / ( √3 + 1 ) }
⇒ x + y = { (√3 + 1)² + (√3 - 1)² } / (√3 - 1) (√3 + 1)
⇒ x + y = ( 3 + 1 +2√3 + 3 + 1 - 2√3 ) / ( 3 - 1 )
⇒ x + y = 8/2
⇒ x + y = 4 --------------------(1)
Now xy = ( √3 - 1 / √3 + 1 ) x ( √3 + 1 / √3 - 1 )
xy = (3 - 1) / (3 - 1) = 2/2 = 1------------------(2)
Given in question,
∴ x²/y + y²/x
= ( x³ + y³ )/xy ------------------- (3)
Apply the algebra formula, A³ + B³ = (A + B)³ - 3AB(A + B)
= (x + y)³ - 3xy(x + y)/xy
Put the value of x + y from (1) and xy from (2) in Equation (3), we will get
= (4)³ - 3 x 1 x (4)/1
= 64 - 3 x 1 x 4
= 64 - 12
= 52