Prove that the sum of the cubes of n real numbers is zero

Question

Today I came across this problem.

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.

I tried to solve it in the following way $$a_1+a_2+\ldots+a_n = 0$$ Squaring both sides $$a_1^2+a_2^2+\ldots+a_n^2+2(0)=0$$ $$a_1^2+a_2^2+\ldots+a_n^2=0$$ However I am unable to move further and get any close to $a_1^3+a_2^3+\ldots+a_n^3=0$. Any help would be appreciated. Also this is a part of an exercise on the topic "No Square is Negative".

As you mentioned , squares are always nonnegative. Hence , a sum of squares can only be $0$ , if all squares are $0$. So, we get $a_i=0$ for all $i$ — Peter, May 23 '23 at 12:58
Indeed, so this answered your question: How Can I prove that $x_1^2+x_2^2+......+x_n^2>0$ by matematical induction — Anne Bauval, May 23 '23 at 13:12
@Henry I used the property that square of the sum of n numbers is equal to the sum of squares of those n numbers and twice their pairwise products to prove the sum of squares = 0. — KKT, May 23 '23 at 13:15
Thanks @Peter for showing me the obvious conclusion which I had missed — KKT, May 23 '23 at 13:16

score 2 · Accepted Answer · answered May 23 '23 at 13:08

2

For every $a_k\in \mathbb{R}$ $a_k^2 \geq 0$ holds.
Hence $\sum_{k=1}^n a_k^2 =0$ implies $a_k=0$ for every $k=1,...,n$. And of course $a_k^3=0^3=0$ follows aswell.

answered May 23 '23 at 13:08

NotABoulderguy

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Prove that the sum of the cubes of n real numbers is zero

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