If $a + b + c = 11$ and $a b + b c + c a = 20$ , then the value of the expression $a^{3} + b^{3} + c^{3} - 3 a b c$ will be

121

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341

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671

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781

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Solution

The correct option is C $671$
${(a^{2} + b^{2} + c^{2})}^{2} = a^{2} + b^{2} + c^{2} + 2 (a b + b c + c a)$
$\Rightarrow a^{2} + b^{2} + c^{2} = 11^{2} - 2 \times 20 = 121 - 40 = 81$
Also, we know that
$(a + b + c) (a^{2} + b^{2} + c^{2} - a b - b c - c a) = a^{3} + b^{3} + c^{3} - 3 a b c$
$\Rightarrow a^{3} + b^{3} + c^{3} - 3 a b c = 11 \times (81 - 20)$
$\Rightarrow a^{3} + b^{3} + c^{3} - 3 a b c = 11 \times 61 = 671$

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