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Expansion of (x+y+z)^2

If a + b + c ...

Question

If a + b + c = 11 and $a^{2} + b^{2} + c^{2} = 81$ , find ; $a b + b c + c a$ .

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Solution

Since $(a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2 (a b + b c + c a) ∴ (11)^{2} = 81 + 2 (a b + b c + c a) ∴ 2 (a b + b c + c a) = 121 - 81 = 40 a b + b c + c a = \frac{40}{2} \Rightarrow a b + b c + c a = 20$

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