If $a + b + c = 8$ and $a b + b c + c a = 20$ , find the value of $a^{3} + b^{3} + c^{3} - 3 a b c$ .

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Solution

The correct option is D 32
Since ${(a + b + c)}^{2} = a^{2} + b^{2} + c^{2} + 2 (a b + b c + c a)$
$a + b + c = 8 and a b + b c + c a = 20$
${(8)}^{2} = a^{2} + b^{2} + c^{2} + 2 \times 20$
$64 = a^{2} + b^{2} + c^{2} + 40$
$a^{2} + b^{2} + c^{2} = 24$
We now use the following identity:
$a^{3} + b^{3} + c^{3} - 3 a b c = (a + b + c) (a^{2} + b^{2} + c^{2} - (a b + b c + c a))$
$a^{3} + b^{3} + c^{3} - 3 a b c = 8 \times (24 - 20) = 4 \times 8 = 32$
Thus, $a^{3} + b^{3} + c^{3} - 3 a b c = 32$

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