If $a + b + C = 12$ and $a^{2} + b^{2} + c^{2} = 50$ , then find $a b + b c + a c$ and $a^{3} + b^{3} + c^{3} - 3 a b c$

800

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400

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700

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36

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Solution

The correct option is C $36$
$a + b + c = 12; a^{2} + b^{2} + c^{2} = 50$
${(a + b + c)}^{2} = 12^{2}$
${(a + b + c)}^{2} = 144$
$a^{2} + b^{2} + c^{2} + 2 (a b + b c + c a) = 144$
$2 (a b + b c + c a) = 144 - 50 = 94$
$a b + b c + c a = 94 / 2 = 47$
$a^{3} + b^{3} + c^{3} - 3 a b c = (a + b + c) (a^{2} + b^{2} + c^{2} - a b - b c - c a)$
$= 12 (50 - (47))$
$= 12 (3)$
$= 36$

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