6. Given a non-constant sequence $\left\{a_{i}\right\}$ satisfies
$$
a_{i}^{2}-a_{i-1} a_{i}+a_{i-1}^{2}=0 \text {, }
$$

and $a_{i+1} \neq a_{i-1}, i=1,2, \cdots, n$.
For a given positive integer $n, a_{1}=a_{n+1}$. Then
$$
\sum_{i=0}^{n-1} a_{i}=
$$
6.0 .

Given $a_{1}^{2}-a_{1-1} a_{i}+a_{1-1}^{2}=0$, so,
$$
a_{i+1}^{2}-a_{i} a_{i+1}+a_{i}^{2}=0 \text {. }
$$

Subtracting the two equations, we get
$$
\left(a_{i+1}-a_{i-1}\right)\left(a_{i+1}+a_{i-1}\right)-a_{i}\left(a_{i+1}-a_{i-1}\right)=0 .
$$

That is, $\left(a_{i+1}-a_{i-1}\right)\left(a_{i+1}+a_{i-1}-a_{i}\right)=0$.
Since $a_{i+1} \neq a_{i-1}$, we have,
$$
a_{i+1}+a_{i-1}-a_{i}=0 .
$$

Therefore, $a_{0}=a_{1}-a_{2}, a_{1}=a_{2}-a_{3}, \cdots$
$$
a_{n-2}=a_{n-1}-a_{n}, a_{n-1}=a_{n}-a_{n+1} .
$$

Thus, $\sum_{i=0}^{n-1} a_{i}=a_{1}-a_{n+1}=0$.