In a triangle ABC,2(bc cos A + ca cos B + ab | vedclass

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(D) Given the expression $2(bc \cos A + ca \cos B + ab \cos C) = 2bc \cos A + 2ca \cos B + 2ab \cos C$. Using the Law of Cosines,we know that $\cos A

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$(a^2 + b^2 + c^2)$

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(D) Given the expression $2(bc \cos A + ca \cos B + ab \cos C) = 2bc \cos A + 2ca \cos B + 2ab \cos C$. Using the Law of Cosines,we know that $\cos A = \frac{b^2 + c^2 - a^2}{2bc}$,$\cos B = \frac{c^2 + a^2 - b^2}{2ca}$,and $\cos C = \frac{a^2 + b^2 - c^2}{2ab}$. Substituting these values into the expression: $= 2bc \left(\frac{b^2 + c^2 - a^2}{2bc}\right) + 2ca \left(\frac{c^2 + a^2 - b^2}{2ca}\right) + 2ab \left(\frac{a^2 + b^2 - c^2}{2ab}\right)$ $= (b^2 + c^2 - a^2) + (c^2 + a^2 - b^2) + (a^2 + b^2 - c^2)$ $= a^2 + b^2 + c^2$. Thus,the value is $a^2 + b^2 + c^2$.

In a $\triangle ABC$,$2(bc \cos A + ca \cos B + ab \cos C) = ?$

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