Do enunciado, temos: $x^3 + y^3 = \underbrace{(x+y)}_{1}(x^2-xy+y^2) = c^2$, logo: $x^2-xy+y^2 = c^2$ e $y = 1-x$ $\implies$ $3x^2 - 3x + 1 = c^2$ . Temos: $x^2 - x\space = \large{\frac{c^2 - 1}{3}}$ $\implies$ $(x-\frac{1}{2})^2 = \large{\frac{c^2 - 1}{3} + \frac{1}{4}}$ $\geq 0$ . Assim: $\large{\frac{c^2 - 1}{3} + \frac{1}{4}}$ $\geq 0$ $\implies$ $4c^2-4+3 \geq 0$ $\implies$ $c^2 \geq \large{\frac{1}{4}}$ . Portanto, $|c| \geq \large{\frac{1}{2}}$ e isso implica que $c \geq \large{\frac{1}{2}}$ ou $c \leq \large{-\space \frac{1}{2}}$. $c \in \space ]-\infty, -\frac{1}{2}]\cup[\frac{1}{2}, +\infty[$ Alternativa $\mathbb{(E)}$