If x=2+\sqrt3,y=2-\sqrt3, then the value of \frac{x^2+y^2}{x^3+y^3} is

RRB Junior Engineer Stage 1 Model Papers 4

Question : 11 of 99

Marks: +1, -0

x=2+√3,y=2-√3

, then the value of

{x^2+y^2}/{x^3+y^3}

Solution:

x=2+√3,y=2-√3

x+y=4,xy=4-3=1

{x^2+y^2}/{x^3+y^3}={(x+y)^2-2xy}/{(x+y)^3-3xy(x+y)}

={16-2}/{64-3×4}=14/52=7/26

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