if x and y are integers such that x ^ 2 = 3 y and xy = 36 , then x – y = ?
here x and y are integers . x ^ 2 = 3 y , xy = 36 . substitute ( x ^ 2 ) / 3 = y in xy = > x ^ 3 = 36 * 3 = > x ^ 3 = 108 . here x ^ 3 is positive , x is also positive . x = 6 then y = 6 . x - y = 0 so option e is correct
subtract(power(multiply(36, 3), const_0_33), divide(36, power(multiply(36, 3), const_0_33)))