Question 3

If $$X = \frac{a^2-(2b-3c)^2}{(3c+a)^2-4b^2} + \frac{4b^2-(3c-a)^2}{(a+2b)^2-9c^2} + \frac{9c^2-(a-2b)^2}{(2b+3c)^2-a^2}$$ 

$$Y = \frac{9y^2-(4z-2x)^2}{(2x+3y)^2-16z^2} + \frac{16z^2-(2x-3y)^2}{(3y+4z)^2-4x^2} + \frac{4x^2-(3y-4z)^2}{(4z+2x)^2-9y^2}$$, 

find $$2017(X+Y)$$.


Correct Answer: 4034

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