problem: Suppose that $x$ and $y$ are real numbers satisfying  \begin{align*}
  4y - 4x^2 &= 1 \\
  4x - 4y^2 &= 1.
  \end{align*} What is $\dfrac{1}{x^3 + y^3}$?
solution: The equations are equivalent to  \begin{align*}
  4x^2 - 4y + 1 &= 0, \\
  4y^2 - 4x + 1 &= 0.
  \end{align*} Summing these equations gives $$4x^2 - 4y + 1 + 4y^2 - 4x + 1 =0,$$ or $$(4x^2 - 4x + 1) + (4y^2 - 4y + 1) = 0.$$ Factoring the squares of binomials gives $$(2x - 1)^2 + (2y-1)^2 = 0.$$ Since squares are always non-negative, it follows that $$2x - 1 = 2y-1 = 0,$$ so $x = y = \frac 12$. The desired answer is $\frac{1}{\frac 18 + \frac 18} = \boxed{4}$.