In triangle ABC,the value of a^3 cos (B-C) + | vedclass

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(D) Using the projection formula,we know $a = b \cos C + c \cos B$,$b = c \cos A + a \cos C$,and $c = a \cos B + b \cos A$. \ Also,by the sine rule,$

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$3abc$

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(D) Using the projection formula,we know $a = b \cos C + c \cos B$,$b = c \cos A + a \cos C$,and $c = a \cos B + b \cos A$. \ Also,by the sine rule,$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$. \ Consider the term $a^3 \cos (B-C)$. Using $\cos (B-C) = \frac{\sin 2B + \sin 2C}{2 \sin (B+C)} = \frac{\sin 2B + \sin 2C}{2 \sin A}$. \ Substituting $a = 2R \sin A$,we get $a^3 \cos (B-C) = (2R \sin A)^2 \cdot a \cdot \frac{\sin 2B + \sin 2C}{2 \sin A} = 2R^2 \sin A (2 \sin B \cos B + 2 \sin C \cos C) \cdot a$. \ This simplifies to $a^2 (b \cos B + c \cos C)$. \ Similarly,$b^3 \cos (C-A) = b^2 (c \cos C + a \cos A)$ and $c^3 \cos (A-B) = c^2 (a \cos A + b \cos B)$. \ Summing these,we get $a^2(b \cos B + c \cos C) + b^2(c \cos C + a \cos A) + c^2(a \cos A + b \cos B)$. \ Rearranging terms: $ab(a \cos B + b \cos A) + bc(b \cos C + c \cos B) + ca(c \cos A + a \cos C)$. \ Using the projection formula $a \cos B + b \cos A = c$,$b \cos C + c \cos B = a$,and $c \cos A + a \cos C = b$,we get: \ $ab(c) + bc(a) + ca(b) = abc + abc + abc = 3abc$.

In $\triangle ABC$,the value of $a^3 \cos (B-C) + b^3 \cos (C-A) + c^3 \cos (A-B)$ is:

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