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Ishaan94

  • 2 years 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.

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

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

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

  4. Ishaan94
    • 2 years ago
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    Thanks :-)

  5. mukushla
    • 2 years ago
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    your welcome

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