Question

A+b+c=8 and a2+b2+c2=64 find ab+ bc+ca

Answer 1

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Heya

Using identity

⏩(a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca⏪

The answer is 0.

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Answer 2

✍✍HERE IS YOUR ANSWER..⬆⬆

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$\underline{SOLUTION}$ ➡

$\underline{Given },\\ \\ a + b + c = 8 \\ \\ a {}^{2} + b {}^{2} + c {}^{2} = 64$


$\underline{To \: find }\: : ab + bc + ca = ?$

$\underline{We \: \: know \: \: that, }\\ \\ \boxed{ (a + b + c) {}^{2} = a {}^{2} + b {}^{2} + c {}^{2} + 2ab + 2bc + 2ca}$

$\underline{From \: \: this \: \: above \: \: identity \: ,\: we \: \: get ,}\\ \\ (a + b + c) {}^{2} = a { }^{2} + b {}^{2} + c {}^{2} + 2ab + 2bc + 2ca \\ \\ = > (a + b + c) {}^{2} = ( a {}^{2} + b {}^{2} + c {}^{2}) + 2( ab + bc + ca ) \\ \\ = > 2(ab + bc + ca) = (a + b + c) {}^{2} -( a {}^{2} + b {}^{2} + c {}^{2} ) \\ \\ = > ab + bc + ca = \frac{(a + b + c) {}^{2} - (a {}^{2} + b {}^{2} + c {}^{2} )}{2} \: \: \: \: \: \: \: - - - - - - > (1)$

$\underline{By \: \: putting \: \: the \: \: values \: \: of \: (a + b + c) {}^{2}} \: \: \\ \underline{and \: \: (a {}^{2} + b {}^{2} + c {}^{2} ) \: \: in \: \: eq.(1) \: ,\: we \: \: get,}$


$ab + bc + ca = \frac{(a + b + c) {}^{2} - (a {}^{2} + b {}^{2} + c {}^{2} }{2} \\ \\ = > ab + bc + ca = \frac{(8) {}^{2} - 64}{2} \\ \\ = > ab + bc + ca = \frac{64 - 64}{2} \\ \\ = > ab + bc + ca = \frac{0}{2} \\ \\ = > ab + bc + ca = 0$



$\underline{ANSWER}$ ➡

$\boxed{\pink{ab + bc + ca = 0}}$

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