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Suppose that \(x\) and \(y\) are real numbers satisfying\(\begin{align*} 4y - 4x^2 &= 1 \\ 4x - 4y^2 &= 1. \end{align*}\)What is \(\dfrac{1}{x^3 + y^3}\)?

 Apr 9, 2020
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4y - 4x2  =  1   --->   y - x2  =  1/4                      (note that y must be positive)

4x - 4y2  =  1   --->   x - y2  =  1/4                      (note that x must be positive)

 

Setting these two equations equal to each other:

      y - x2  =  x - y2

     y2 + y  =  x2 + x

Completing the square:

    y2 + y + 1/4  =  x2 + x + 1/4

        (y + 1/2)2  =  (x + 1/2)2 

Taking the square root of both sides (ignoring negative possibilities):

        y + 1/2  =  x + 1/2

                 y  =  x

 

Since  4x - 4y2  =  1   --->   4x - 4x2  =  1   --->   x2 - x - 1/4  =  0

                                  --->   (x - 1/2)2  =  0   --->   x  =  1/2     and     y  =  1/2

 

x+ y3  =  (1/2)3 + (1/2)3  =  1/4

 

1 / (1/4)  =  4

 Apr 9, 2020

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