Let $x, y$ be real numbers satisfying
\[
\left(x^{2}+x-1\right)\left(x^{2}-x+1\right)=2\left(y^{3}-2 \sqrt{5}-1\right)
\]
and
\[
\left(y^{2}+y-1\right)\left(y^{2}-y+1\right)=2\left(x^{3}+2 \sqrt{5}-1\right)
\]
Find $8 x^{2}+4 y^{3}$.

Step 1 ：Given the equations, we can rewrite them as follows: \(x^4 - 1 = 2(y^3 - 2\sqrt{5} - 1)\) and \(y^4 - 1 = 2(x^3 + 2\sqrt{5} - 1)\)

Step 2 ：Adding these two equations together, we get: \(x^4 + y^4 - 2 = 2(y^3 - x^3)\)

Step 3 ：Rearranging the equation, we get: \(x^4 - 2y^3 + y^4 - 2x^3 = 2\)

Step 4 ：Multiplying the equation by 4, we get: \(4x^4 - 8y^3 + 4y^4 - 8x^3 = 8\)

Step 5 ：Rearranging the equation, we get: \(4x^4 - 8x^3 + 4y^4 - 8y^3 = 8\)

Step 6 ：Grouping the terms, we get: \(4(x^4 - 2x^3) + 4(y^4 - 2y^3) = 8\)

Step 7 ：Factoring out 4, we get: \(4[(x^4 - 2x^3) + (y^4 - 2y^3)] = 8\)

Step 8 ：Dividing both sides by 4, we get: \((x^4 - 2x^3) + (y^4 - 2y^3) = 2\)

Step 9 ：Rewriting the equation in terms of the expression we are asked to find, we get: \(8x^2 + 4y^3 = 2\)

Step 10 ：Therefore, the final answer is \(\boxed{8x^2 + 4y^3 = 2}\)