If x = 12 and y = 7, then the value of $$\frac{(x^2+y^2-xy)}{(x^3+y^3 )}$$​ is: 

Correct option is B

Given: ​

​$$\frac{(x^2+y^2-xy)}{(x^3+y^3 )}$$   

where x = 12, y = 7 

Formula Used: ​

​$$(a+b)^3 = (a+b)(a^2+b^2-ab)$$ 

Solution: 

​$$\frac{(x^2+y^2-xy)}{(x^3+y^3 )}$$ 

By using the formula 

​$$\begin{aligned}&=\frac{(x^2+y^2-xy)}{(x^3+y^3 )} \\&=\frac{(x^2+y^2-xy)}{(x + y)(x^2 + y^2 - xy)} \\& =\frac{ 1}{x+y} \end{aligned}$$     

Now, putting the value of x and y 

​$$\frac{1}{x+y} = \frac{1}{12+7} = \frac{1}{19}$$     

thus, option(b) is correct.