Solution

As we know, (x³ + y³) = (x + y) (x² + y² - xy) (x³ – y³) = (x - y) (x² + y² + xy) (x + y)² - (x - y)² = 4xy [(x - y)² + 3xy] = [x² + y² - 2xy + 3xy] = [x² + y² + xy] [(x + y)² - 3xy] = [x² + y² + 2xy - 3xy] = [x² + y² - xy] Now- (x³-y³)/x[(x+y)²-3xy] ÷y(x-y)²+3xy]/ x³+y³ ×(x+y)²-(x-y)² /x²- y² = (x - y) (x² + y² + xy)/x [x² + y² - xy] ÷ y [x² + y² + xy]/ (x + y) (x² + y² - xy) × 4xy/(x-y) (x+y) = (x - y) (x² + y² + xy)/x [x² + y² - xy] × (x + y) (x² + y² - xy)/ y [x² + y² + xy] × 4xy/(x-y) (x+y) =4