Q) The sum of the reciprocals of two real numbers is −1, and the sum of their cubes is 4. What are the numbers?
A)
1/x + 1/y = -1
x+y / xy = -1
x+y = -xy
[and (x+y)^2 = x^2.y^2 ]
Now x^3 + y^3 = 4
(x+y)(x^2 - xy + y^2) = 4
(-xy)(x^2.y^2 - 2xy - xy)
(xy)(3xy - x^2.y^2) = 4
(xy)(xy)(3-xy) = 4
Thus, xy = -1
x = -1/y
1/x + 1/y = -1
-y + 1/y = -1
y = (1 +- rt5) / 2
and x= -1/y
so you get the corresponding values of x.
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Monday, August 11, 2008
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