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If $$x \ne -1$$, $$y \ne -1$$ and $$x^3 + y^3 + 3xy = 1$$ then
Write the equation as $$x^3 + y^3 + (-1)^3 - 3xy(-1) = 0$$, which factorises as $$(x + y - 1) \cdot \frac{1}{2}\left[(x - y)^2 + (y + 1)^2 + (x + 1)^2\right] = 0$$. The bracket is zero only when $$x = y = -1$$, which is excluded, so $$x + y = 1$$. Then $$(x + y)^2 = 1$$ also holds, so options a and d are both correct.
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