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Ishaan94
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The sum of \(n\) real numbers is zero and the sum of their pairwise products is also zero. Prove that the sum of the cubes of the numbers is zero.
 2 years ago
 2 years ago
Ishaan94 Group Title
The sum of \(n\) real numbers is zero and the sum of their pairwise products is also zero. Prove that the sum of the cubes of the numbers is zero.
 2 years ago
 2 years ago

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mukushla Group TitleBest ResponseYou've already chosen the best response.2
\[(a _{1}+a _{2}+a _{3}+...+a _{n})^2=a _{1}^2+a _{2}^2+a _{3}^2+...+a _{n}^2+2(a _{1}a _{2}+a _{1}a _{3}+...+a _{n1}a _{n})\\ a _{1}^2+a _{2}^2+a _{3}^2+...+a _{n}^2=(a _{1}+a _{2}+a _{3}+...+a _{n})2(a _{1}a _{2}+a _{1}a _{3}+...+a _{n1}a _{n})=0 \]
 2 years ago

mukushla Group TitleBest ResponseYou've already chosen the best response.2
then \[a_{1}=a_{2}=a_{3}=...=a_{n}=0 \\ so \\ \] sum of the cubes of the numbers is zero
 2 years ago

Ishaan94 Group TitleBest ResponseYou've already chosen the best response.0
I see. This wasn't that hard. Only my brain stopped thinking.
 2 years ago

Ishaan94 Group TitleBest ResponseYou've already chosen the best response.0
Thanks :)
 2 years ago

mukushla Group TitleBest ResponseYou've already chosen the best response.2
your welcome
 2 years ago
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