This is a corrected version.
Let $a_1,a_2,a_3,b_1,b_2,b_3,b_4,b_5,b_6\in \mathbb{C}$ such that $a_i\not=a_j$ for all $i\not=j.$
If $$\begin{vmatrix} a_1 & a_2& a_3 & b_1 \\ a_1^2 & a_2^{2} & a_3^{2} & b_2\\ a_1^3 & a_2^{3} & a_3^{3} & b_3\\ a_1^4 & a_2^{4} & a_3^{4} & b_4\\ \end{vmatrix} =0,$$ $$\begin{vmatrix} a_1^2 & a_2^{2} & a_3^{2} & b_2\\ a_1^3 & a_2^{3} & a_3^{3} & b_3\\ a_1^4 & a_2^{4} & a_3^{4} & b_4\\ a_1^5 & a_2^{5} & a_3^{5} & b_5\\ \end{vmatrix} =0,$$ and $$\begin{vmatrix} a_1^3 & a_2^{3} & a_3^{3} & b_3\\ a_1^4 & a_2^{4} & a_3^{4} & b_4\\ a_1^5 & a_2^{5} & a_3^{5} & b_5\\ a_1^6 & a_2^{6} & a_3^{6} & b_6\\ \end{vmatrix} =0,$$ then all minors of order $4$ of the matrix
$$\begin{bmatrix} a_1 & a_2& a_3 & b_1 \\ a_1^2 & a_2^{2} & a_3^{2} & b_2\\ a_1^3 & a_2^{3} & a_3^{3} & b_3\\ a_1^4 & a_2^{4} & a_3^{4} & b_4\\ a_1^5 & a_2^{5} & a_3^{5} & b_5\\ a_1^6 & a_2^{6} & a_3^{6} & b_6\\ \end{bmatrix}$$ are $0$. It is stated in a paper that this is true without proof. I believe that it is related with Vandermonde determinant but I do not know how to prove it. Could you please help me or give me an idea? Thank you so much for your help.
Masik
Let \begin{eqnarray}v_i=(a_1^i, a_2^i, a_3^i, b_i)\end{eqnarray} That the three determinants are 0 implies that the row vectors are linearly dependent. In particular, $\text{dim Span}\{v_1, v_2, v_3, v_4\}\leq 3$, $\text{dim Span}\{v_2, v_3, v_4, v_5\}\leq 3$, $\text{dim Span}\{v_3, v_4, v_5, v_6\}\leq 3$. So \begin{eqnarray}\text{dim Span}\{v_1, v_2, v_3, v_4, v_5\}\leq \text{dim Span}\{v_1, v_2, v_3, v_4\}+\text{dim Span}\{v_2, v_3, v_4, v_5\}-3\leq 3\end{eqnarray} Similarly, \begin{eqnarray}\text{dim Span}\{v_1, v_2, v_3, v_4, v_5, v_6\}\leq\text{dim Span}\{v_1, v_2, v_3, v_4, v_5\}+\text{dim Span}\{v_3, v_4, v_5, v_6\}-3\leq 3\end{eqnarray}
So any 4 row vectors are linearly dependent and the determinant of any $4\times 4$ minor is 0.