1. The problem asks to express the transformed variables $X$ and $Y$ in terms of the original variables $x$ and $y$ from the given function $$y = \frac{2x}{x^3 - 8}.$$\n\n2. To linearize this function, we look for substitutions $X$ and $Y$ such that the relationship between $X$ and $Y$ is linear (a straight line).\n\n3. Notice that the denominator can be factored as a difference of cubes: $$x^3 - 8 = (x - 2)(x^2 + 2x + 4).$$ However, the key is to rewrite the equation to isolate terms for linearization.\n\n4. Multiply both sides by the denominator: $$y(x^3 - 8) = 2x.$$\n\n5. Rearranged: $$y x^3 - 8 y = 2 x.$$\n\n6. To get a linear form, define new variables: $$X = x^3 - 8$$ and $$Y = \frac{2x}{y}.$$\n\n7. From the original equation, substituting $X$ and $Y$ gives: $$y = \frac{2x}{x^3 - 8} \implies y X = 2x \implies Y = y X = 2x,$$ but since $Y = \frac{2x}{y}$, this suggests a slight correction. Instead, rearranging the original equation as $$y = \frac{2x}{x^3 - 8} \implies y (x^3 - 8) = 2x,$$ so if we let $$Y = y$$ and $$X = x^3 - 8,$$ then the equation becomes $$Y X = 2x,$$ which is not linear in $X$ and $Y$.\n\n8. Alternatively, define $$X = x^3 - 8$$ and $$Y = \frac{2x}{y}.$$ Then from the original equation, $$y = \frac{2x}{x^3 - 8} \implies y = \frac{2x}{X} \implies y X = 2x \implies Y = \frac{2x}{y} = X.$$\n\n9. This means $$Y = X,$$ which is a linear equation representing a straight line with slope 1.\n\n10. Therefore, the linearized variables are: $$X = x^3 - 8$$ and $$Y = \frac{2x}{y}.$$\n\nFinal answer:\n$$X = x^3 - 8, \quad Y = \frac{2x}{y}.$$