If $a + b + c = 6, a^{2} + b^{2} + c^{2} = 14 and a^{3} + b^{3} + c^{3} = 36$ then find the value of $a b c$

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Solution

$a + b + c = 6, a^{2} + b^{2} + c^{2} = 14, a^{3} + b^{3} + c^{3} = 36$
$(a + b + c) = a^{2} + b^{2} + c^{2} + 2 (a b + b c + c a)$
$⟹ 2 (a b + b c + c a) = (6)^{2} - 14$
$⟹ 2 (a b + b c + c a) = (6)^{2} - 14$
$= 36 - 14 = 22$
$a b + b c + c a = 11$
Now ,we know $a^{3} + b^{3} + c^{3} - 3 a b c = (a + b + c) (a^{2} + b^{2} + c^{2} - (a b + b c + c a))$
$⟹ 36 - 3 a b c = 6 (14 - 11)$
$⟹ 36 - 3 a b c = 6 \times 3 = 18$
$⟹ 3 a b c = 18$
$⟹ a b c = 6$

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