In Delta ABC,{a^2}({cos ^2}B - {cos ^2}C) + { | vedclass

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(A) Using the sine rule,we have $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$,so $a = 2R \sin A$,$b = 2R \sin B$,and $c = 2R \sin C$.S

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(A) Using the sine rule,we have $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$,so $a = 2R \sin A$,$b = 2R \sin B$,and $c = 2R \sin C$.Substituting these into the expression:$E = \sum a^2(\cos^2 B - \cos^2 C) = \sum (2R \sin A)^2(\cos^2 B - \cos^2 C)$Since $\cos^2 	heta = 1 - \sin^2 	heta$,we have $\cos^2 B - \cos^2 C = (1 - \sin^2 B) - (1 - \sin^2 C) = \sin^2 C - \sin^2 B$.Thus,$E = 4R^2 \sum \sin^2 A(\sin^2 C - \sin^2 B)$.Using $a = 2R \sin A$,we have $\sin A = \frac{a}{2R}$,so $\sin^2 A = \frac{a^2}{4R^2}$.$E = 4R^2 \left[ \frac{a^2}{4R^2}(\sin^2 C - \sin^2 B) + \frac{b^2}{4R^2}(\sin^2 A - \sin^2 C) + \frac{c^2}{4R^2}(\sin^2 B - \sin^2 A) ight]$$E = a^2(\sin^2 C - \sin^2 B) + b^2(\sin^2 A - \sin^2 C) + c^2(\sin^2 B - \sin^2 A)$Substituting $\sin^2 A = \frac{a^2}{4R^2}$,$\sin^2 B = \frac{b^2}{4R^2}$,$\sin^2 C = \frac{c^2}{4R^2}$:$E = a^2(\frac{c^2}{4R^2} - \frac{b^2}{4R^2}) + b^2(\frac{a^2}{4R^2} - \frac{c^2}{4R^2}) + c^2(\frac{b^2}{4R^2} - \frac{a^2}{4R^2})$$E = \frac{1}{4R^2} [a^2c^2 - a^2b^2 + b^2a^2 - b^2c^2 + c^2b^2 - c^2a^2] = 0$.

In $\Delta ABC$,${a^2}({\cos ^2}B - {\cos ^2}C) + {b^2}({\cos ^2}C - {\cos ^2}A) + {c^2}({\cos ^2}A - {\cos ^2}B) = $

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