linear algebra - Vandermonde determinant and linearly independent

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