In $\Delta ABC$, $b^2 \cos 2A - a^2 \cos 2B =$

Using the cosine rule, $$\cos A=\frac{b^2+c^2-a^2}{2bc}$$ and $$\cos B=\frac{a^2+c^2-b^2}{2ac}.$$ Now use $$\cos 2A=2\cos^2 A-1$$ and $$\cos 2B=2\cos^2 B-1.$$ Then $$b^2\cos 2A-a^2\cos 2B=2b^2\cos^2 A-2a^2\cos^2 B-(b^2-a^2).$$ Substituting the values of the cosines, $$2b^2\cos^2 A=\frac{(b^2+c^2-a^2)^2}{2c^2}$$ and $$2a^2\cos^2 B=\frac{(a^2+c^2-b^2)^2}{2c^2}.$$ Therefore, $$b^2\cos 2A-a^2\cos 2B=\frac{(b^2+c^2-a^2)^2-(a^2+c^2-b^2)^2}{2c^2}-(b^2-a^2).$$ Using difference of squares, $$\frac{[(b^2+c^2-a^2)-(a^2+c^2-b^2)][(b^2+c^2-a^2)+(a^2+c^2-b^2)]}{2c^2}-(b^2-a^2).$$ This becomes $$\frac{2(b^2-a^2)\cdot 2c^2}{2c^2}-(b^2-a^2)=2(b^2-a^2)-(b^2-a^2).$$ Hence $$b^2\cos 2A-a^2\cos 2B=b^2-a^2.$$ Comparing with the options, the correct choice is $A$.