If $a + b + c = 12$ and $a^{2} + b^{2} + c^{2} = 66$ . Find the value of $44 a b + b c + c a$

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Solution

We know that $(a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2 a b + 2 b c + 2 c a$
i.e. $(a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2 (a b + b c + c a)$
Substituting, the given values we get,
$12^{2} = 66 + 2 (a b + b c + c a)$
$\Rightarrow 144 - 66 = 2 (a b + b c + c a)$
$\Rightarrow 78 = 2 (a b + b c + c a)$
$\Rightarrow a b + b c + c a = \frac{78}{2} = 39$

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