Take three real numbers $a, b$ and $c$ such that the roots of equation $x^3+ax^2+bx+c=0$ have the same absolute value. We need to show that $a=0$ if and only if $b=0$.
I tried taking the roots as $p, p, p$ or $p, -p, p$ and so on and showing with Vieta's formulae how $p=0$ and hence the sum of roots, which in this case, should be $-a$, is equal to zero, so $a=0$. I am looking for a better hint on how to proceed.
The problem given is equivalent to the following situation:
Given $|p|=|q|=|r|$, prove that $p+q+r=0$ if and only if $pq+qr+rp=0$.
($\rightarrow$) As there are only three variables, the number of positive and negative numbers must differ by an odd number. Hence if $p+q+r=0$, this implies that $p=q=r=0$.
($\leftarrow$) Again, the number of positive terms and negative terms in $pq, qr, rp$ must differ by an odd number(You can list them all, i.e when all are positive, when only one is positive,...) Hence $p=q=r=0$.
Hence we are done. This proof is essentially same as what you thought, but it is more concise.
I'm proving what @Joshua Woo had written. I think you'll be able to continue from there.
$$pq+qr+rp=0$$ then $$2pq+2qr+2rp=0$$ This means that $$(p+q+r)^2=p^2+q^2+r^2+2pq+2qr+2rp=p^2+q^2+r^2$$ Hence $(p+q+r)$ can only be equal to zero if $p=0,q=0,r=0$ and $2pq+2qr+2pr=0$, given that $p,q,r$ are real
