If $a + b + c = 8$ and $a^{2} + b^{2} + c^{2} = 40$ . Find $a b + b c + c a$ .

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Solution

Given that,
$a + b + c = 8$ and $a^{2} + b^{2} + c^{2} = 40$

we know that,

$(a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2 (a b + b c + c a)$

$=> 2 (a b + b c + c a) = (a + b + c)^{2} - (a^{2} + b^{2} + c^{2})$

$=> (a b + b c + c a) = \frac{(a + b + c)^{2} - (a^{2} + b^{2} + c^{2})}{2}$

$=> \frac{8^{2} - 40}{2}$
$=> \frac{64 - 40}{2} = 12$

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Similar questions

If a + b + c = 9 and ab + bc + ca = 40, find a² + b² + c².

\frac{a^{2} + b^{2} + a b}{b^{2} + c^{2} + b c} + \frac{c^{2} + c a + a^{2}}{b^{2} + c^{2} + b c}

a^{2} + b^{2} + c^{2} = 20

a + b + c = 0

a b + b c + c a

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