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Question 121
If $$x = 2 + \sqrt{3}, y = 2 - \sqrt{3}$$, then the valuer of $$\frac{x^2 + y^2}{x^3 + y^3}$$ isÂ
Solution
$$\frac{x^2 + y^2}{x^3 + y^3}$$
 or $$\frac{(x + y)^2 - 2xy}{(x + y)^3 - 3xy(x+y)}$$
as x+y = 4
and xy = 1
Now after putting values of x+y and xy and solving, we will get its value as 7/26.
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