If $$x$$, $$y$$ are real numbers such that $$x^2 + y + 8 = 6x^{2/3}y^{1/3}$$, $$x^{2/3} + y^{1/3} > -2$$ then

Solution

Put $$u = x^{2/3}$$, $$v = y^{1/3}$$, $$w = 2$$; the equation is $$u^3 + v^3 + w^3 - 3uvw = 0$$, i.e. $$(u + v + w)\left[(u - v)^2 + (v - w)^2 + (w - u)^2\right] = 0$$. Since $$u + v + 2 > 0$$, we must have $$u = v = w = 2$$, so $$x^{2/3} = 2$$ and $$y = 8$$. Then $$x^2 = 8$$, so $$x = \pm 2\sqrt{2}$$ and option d need not hold. Hence options a and b are both correct.

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