If x is a rational number and $$\frac{(x+1)^3-(x-1)^3}{(x+1)^2-(x-1)^2}=2$$ then the sum of numerator and denominator of x is
Given , $$\frac{(x+1)^3-(x-1)^3}{(x+1)^2-(x-1)^2}=2$$
On simplifying , we get $$\frac{6x^2 +2}{4x}=2$$
$$\frac{3x^2 +1}{2x}=2$$
$$3x^2 +1=4x$$
$$3x^2-4x+1=0$$
(3x-1)(x-1)=0
As it was mentioned that x is rational number . Hence, x= $$\frac{1}{3}$$
Hence , 1+3 = 4.
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