If x=100,y=99 then find the value of $$(x^2+y^2+xy)/(x^3-y^3 ) .$$​​

Correct option is A

Given:
x = 100
y = 99 
Formula Used:
​$$x^3 - y^3 = (x - y)(x^2 + xy + y^2)$$​​
Solution:
$$\frac{x^2 + y^2 + xy}{x^3 - y^3}$$​ = $$\frac{x^2 + y^2 + xy}{(x - y)(x^2 + xy + y^2)}$$​ = $$\frac{1}{(x - y)}$$​ 
Putting the value of  x and } y
​$$= \frac{1}{(100 - 99)}$$​​
​$$= \frac{1}{1}$$​​
= 1