I had to show that $$\iiint_{R^3} Y\cdot(\nabla\times X)dV=\iiint_{R^3} X\cdot (\nabla \times Y)dV$$ By using the identity as indicated by the excercise $$Y\cdot(\nabla\times X)-X\cdot (\nabla \times Y)=\nabla\cdot(X\times Y)$$ I get
$$\iiint_{R^3} \nabla\cdot(X\times Y)\ dV =0$$ Does anyone know how can I prove that? Maybe you can use the divergence theorem and turn it into surface integral but I have been trying with no success. Any help would be really appreciated.