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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.
 one year ago
 one year ago
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.
 one year ago
 one year ago

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

mukushlaBest 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
 one year ago

Ishaan94Best ResponseYou've already chosen the best response.0
I see. This wasn't that hard. Only my brain stopped thinking.
 one year ago
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