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

We know that one of the identity is $(a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2 (a b + b c + a c)$ and we are given that $a + b + c = 9$ and $a b + b c + a c = 26$ , therefore, we have:

$(a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2 (a b + b c + a c) \Rightarrow 9^{2} = a^{2} + b^{2} + c^{2} + (2 \times 26) \Rightarrow 81 = a^{2} + b^{2} + c^{2} + 52 \Rightarrow a^{2} + b^{2} + c^{2} = 81 - 52 \Rightarrow a^{2} + b^{2} + c^{2} = 29$

Hence, $a^{2} + b^{2} + c^{2} = 29$

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