In any triangle $A B C$ , the value of $a (b^{2} + c^{2}) cos A + b (c^{2} + a^{2}) cos B + c (a^{2} + b^{2}) cos C$ is

{3 a b c}^{2}

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{3 a}^{2} b c

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3 a b c

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{3 a b}^{2} c

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Solution

The correct option is C $3 a b c$
To find : $a (b^{2} + c^{2}) c o s A + b (c^{2} + a^{2}) c o s B + C (a^{2} + b^{2}) c o s C$
$a (b^{2} + c^{2}) c o s A + b (c^{2} + a^{2}) c o s B + C (a^{2} + b^{2}) c o s C$
$= a b^{2} c o s A + a c^{2} c o s A + b c^{2} c o s B + b a^{2} c o s B +$
$c a^{2} c o s C + c b^{2} c o s C$
$= a b^{2} c o s A + b a^{2} c o s B + a c^{2} c o s A + c a^{2} c o s C$
$+ b c^{2} c o s B + c b^{2} c o s C$
$= a b (b c o s A + a c o s B) + a c (C c o s A + a c o s C)$
$+ b c (C c o s B + b c o s C)$
$= a b c + a b c + a b c$
(By projection formula)
$= 3 a b c$
Ans : $3 a b c$

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