With usual notations, in triangle ABC, if 2a^ | vedclass

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(B) Given $2a^2 = b^2 + c^2$. By the Law of Cosines, $\cos A = \frac{b^2 + c^2 - a^2}{2bc}$.Substituting $b^2 + c^2 = 2a^2$, we get $\cos A = \frac{2a

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$(\frac{b^2 - c^2}{2bc})^2$

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(B) Given $2a^2 = b^2 + c^2$. By the Law of Cosines, $\cos A = \frac{b^2 + c^2 - a^2}{2bc}$.Substituting $b^2 + c^2 = 2a^2$, we get $\cos A = \frac{2a^2 - a^2}{2bc} = \frac{a^2}{2bc}$.Using the identity $\cos 3A = 4\cos^3 A - 3\cos A$, we have $\frac{\cos 3A}{\cos A} = 4\cos^2 A - 3$.Thus, $\frac{\cos 3A}{\cos A} + 2 = 4\cos^2 A - 3 + 2 = 4\cos^2 A - 1$.Substitute $\cos A = \frac{b^2 + c^2 - a^2}{2bc}$. Since $a^2 = \frac{b^2 + c^2}{2}$, $\cos A = \frac{b^2 + c^2 - (b^2 + c^2)/2}{2bc} = \frac{b^2 + c^2}{4bc}$.Then $4\cos^2 A - 1 = 4(\frac{b^2 + c^2}{4bc})^2 - 1 = \frac{(b^2 + c^2)^2}{4b^2c^2} - 1 = \frac{b^4 + c^4 + 2b^2c^2 - 4b^2c^2}{4b^2c^2} = \frac{b^4 + c^4 - 2b^2c^2}{4b^2c^2} = \frac{(b^2 - c^2)^2}{4b^2c^2} = (\frac{b^2 - c^2}{2bc})^2$.

With usual notations, in $\triangle ABC$, if $2a^2 = b^2 + c^2$, then $\frac{\cos 3A}{\cos A} + 2 = $

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