If $a + b + c = 6$ and $a b + b c + c a = 11$ , then the value of $b c (b + c) + c a (c + a) + a b (a + b) + 3 a b c$ is

33

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66

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55

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32

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Solution

The correct option is B $66$
$b c (b + c) + c a (c + a) + a b (a + b) + 3 a b c$
$= b^{2} c + c^{2} b + c^{2} a + a^{2} b + a^{2} b + b^{2} a + a b c + a b c + a b c$
$= b c (a + b + c) + a c (a + b + c) + a b (a + b + c)$
$= (a + b + c) (a b + b c + a c) = 6 \times 11 = 6$
Alternate Method : If $a = 1, b = 2$ and $c = 3$ then,
$b c (b + c) + c a (c + a) + a b (a + b) + 3 a b c$
$= 6 \times 5 + 3 \times 4 + 2 \times 3 + 3 \times 6 = 30 + 12 + 6 + 18 = 66$ .

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