Note that we can write
1 1
______ - x^3 - y^3 as ________ - ( x ^3 + y^3)
x^3y^3 x^3y^3
xy ( x + y) = 1 ⇒ x^2y + y^2x = 1 ⇒ [ 3x^2y + 3y^2x ] = 3
x + y = 1 / [xy] cube both sides
x^3 + 3x^2y + 3y^2x + y^3 = 1 / [x^3y^3 ]
x^3 + [ 3x^2y + 3y^2x ] + y^3 = 1 / {x^3y^3]
x^3 + 3 + y^3 = 1 / [ x^3y^3]
3 = 1 / [ x^3y^3 ] - (x^3 + y^3]