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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