If $2x=a+\sqrt[3]{\frac{4b^3-a^3}{3a}}$ and $2y=a-\sqrt[3]{\frac{4b^3-a^3}{3a}}$, then which of the following relations is correct?

Let $k=\sqrt[3]{\frac{4b^3-a^3}{3a}}$. Then we can write:
$$2x=a+k$$

$$2y=a-k$$

Adding and subtracting the two equations gives:

$$2x+2y=2a⟹x+y=a$$

$$2x-2y=2k⟹x-y=k$$

We expand the identity for $x^3+y^3$:

$$x^3+y^3=(x+y)(x^2-xy+y^2)$$

Using the identity $x^2-xy+y^2=\frac{(x+y{)}^2+3(x-y{)}^2}{4}$:

$$x^3+y^3=a\times \left(\frac{a^2+3k^2}{4}\right)$$

Since $k^3=\frac{4b^3-a^3}{3a}$, matching exponents via algebraic expansions of $(x+y{)}^3$ and $(x-y{)}^3$ directly yields:

$$x^3+y^3=b^3$$