RRB JE 29th May 2019 Shift-3

Question 52

If $$x = 100, y = 99$$ then find the value of $$\frac{x^2  + y^2 + xy}{x^3 - y^3}$$

Solution

$$(x^3-y^3)=(x-y)(x^2+xy+y^2)$$

So, required ratio= 1/(x-y)=1/(100-99)=1.

D is correct choice.

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