Find $a^{2} + b^{2} + c^{2}; i f a + b + c = 9 a n d a b + b c + c a = 24$

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Solution

$a + b + c = 9 \Rightarrow (a + b + c)^{2} = (9)^{2} \Rightarrow a^{2} + b^{2} + c^{2} + 2 a b + 2 b c + 2 c a = 81 \Rightarrow a^{2} + b^{2} + c^{2} + 2 (a b + b c + c a) = 81 \Rightarrow a^{2} + b^{2} + c^{2} + 2 \times 24 = 81 \Rightarrow a^{2} + b^{2} + c^{2} + 48 = 81 \Rightarrow a^{2} + b^{2} + c^{2} = 81 - 48 \Rightarrow a^{2} + b^{2} + c^{2} = 33$ .

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If $a + b + c = 9$ and $a b + b c + c a = 40$ . Find the value of $a^{2} + b^{2} + c^{2}$ .

\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}

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