In a triangle ABC, {a^3}cos (B - C) + {b^3}co | vedclass

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(B) Using the sine rule,$a = k \sin A, b = k \sin B, c = k \sin C.$Substituting these into the expression:$E = k^3 [\sin^3 A \cos(B-C) + \sin^3 B \cos

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$3abc$

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(B) Using the sine rule,$a = k \sin A, b = k \sin B, c = k \sin C.$Substituting these into the expression:$E = k^3 [\sin^3 A \cos(B-C) + \sin^3 B \cos(C-A) + \sin^3 C \cos(A-B)]$Since $A+B+C = \pi,$ we have $\sin A = \sin(B+C).$Thus,$\sin^3 A \cos(B-C) = \sin^2 A \sin(B+C) \cos(B-C) = \sin^2 A \cdot \frac{1}{2} (\sin 2B + \sin 2C).$Expanding this for all terms and simplifying using the identity $\sin A \sin B \sin C = \frac{abc}{k^3},$ we get:$E = 3k^3 \sin A \sin B \sin C = 3abc.$

In a triangle $ABC,$ ${a^3}\cos (B - C) + {b^3}\cos (C - A) + {c^3}\cos (A - B) = $

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