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

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(D) Using the projection formula,we know that $a = b \cos C + c \cos B$,$b = c \cos A + a \cos C$,and $c = a \cos B + b \cos A$.We can write $a^3 \cos

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$3abc$

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(D) Using the projection formula,we know that $a = b \cos C + c \cos B$,$b = c \cos A + a \cos C$,and $c = a \cos B + b \cos A$.We can write $a^3 \cos(B-C) = a^2 \cdot a \cos(B-C)$.Since $a = 2R \sin A = 2R \sin(180^\circ - (B+C)) = 2R \sin(B+C)$,we have:$a^3 \cos(B-C) = a^2 \cdot 2R \sin(B+C) \cos(B-C) = a^2 R [\sin(2B) + \sin(2C)] = a^2 R [2 \sin B \cos B + 2 \sin C \cos C]$.Using $b = 2R \sin B$ and $c = 2R \sin C$,this becomes:$a^3 \cos(B-C) = a^2 (b \cos B + c \cos C) = a^2 b \cos B + a^2 c \cos C \quad \dots (i)$.Similarly,$b^3 \cos(C-A) = b^2 c \cos C + b^2 a \cos A \quad \dots (ii)$$c^3 \cos(A-B) = c^2 a \cos A + c^2 b \cos B \quad \dots (iii)$Adding $(i), (ii),$ and $(iii)$:Sum $= (a^2 b \cos B + b^2 a \cos A) + (b^2 c \cos C + c^2 b \cos B) + (c^2 a \cos A + a^2 c \cos C)$$= ab(a \cos B + b \cos A) + bc(b \cos C + c \cos B) + ca(c \cos A + a \cos C)$$= ab(c) + bc(a) + ca(b) = abc + abc + abc = 3abc$.

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

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