Here's the question you clicked on:
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.
\[(a _{1}+a _{2}+a _{3}+...+a _{n})^2=a _{1}^2+a _{2}^2+a _{3}^2+...+a _{n}^2+2(a _{1}a _{2}+a _{1}a _{3}+...+a _{n-1}a _{n})\\ a _{1}^2+a _{2}^2+a _{3}^2+...+a _{n}^2=(a _{1}+a _{2}+a _{3}+...+a _{n})-2(a _{1}a _{2}+a _{1}a _{3}+...+a _{n-1}a _{n})=0 \]
then \[a_{1}=a_{2}=a_{3}=...=a_{n}=0 \\ so \\ \] sum of the cubes of the numbers is zero
I see. This wasn't that hard. Only my brain stopped thinking.