If $a + b + c = 9$ and $a b + b c + a c = 26$ . Find $a^{2} + b^{2} + c^{2}$ .

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Solution

Given that,
$a + b + c = 9$ and
$a b + b c + c a = 26$

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

$(9)^{2} = a^{2} + b^{2} + c^{2} + 2 (26)$

$81 = a^{2} + b^{2} + c^{2} + 52$

$a^{2} + b^{2} + c^{2} = 81 - 52 = 29$

$∴ a^{2} + b^{2} + c^{2} = 29.$

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