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

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 $\boxed{{\displaystyle 8x^2+4y^3=2}}$