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Ishaan94

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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  1. mukushla
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    \[(a _{1}+a _{2}+a _{3}+...+a _{n})^2=a _{1}^2+a _{2}^2+a _{3}^2+...+a _{n}^2+2(a _{1}a _{2}+a _{1}a _{3}+...+a _{n-1}a _{n})\\ a _{1}^2+a _{2}^2+a _{3}^2+...+a _{n}^2=(a _{1}+a _{2}+a _{3}+...+a _{n})-2(a _{1}a _{2}+a _{1}a _{3}+...+a _{n-1}a _{n})=0 \]

    • one year ago
  2. mukushla
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    then \[a_{1}=a_{2}=a_{3}=...=a_{n}=0 \\ so \\ \] sum of the cubes of the numbers is zero

    • one year ago
  3. Ishaan94
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    I see. This wasn't that hard. Only my brain stopped thinking.

    • one year ago
  4. Ishaan94
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    Thanks :-)

    • one year ago
  5. mukushla
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    your welcome

    • one year ago
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