Suppose that \(x\) and \(y\) are real numbers satisfying\(\begin{align*} 4y - 4x^2 &= 1 \\ 4x - 4y^2 &= 1. \end{align*}\)What is \(\dfrac{1}{x^3 + y^3}\)?
+23246 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
x3 + y3 = (1/2)3 + (1/2)3 = 1/4
1 / (1/4) = 4
