Why is there another transformation matrix for the bases of the image and of the preimage of this mapping? There is the following Matrix:
\begin{pmatrix}1&1&1\\ a&b&c\\ a^2&b^2&c^2\end{pmatrix}
At a point it is needed to calculate the determinant of the matrix. In the official solution it is written:
$det\begin{pmatrix}1&1&1\\ a&b&c\\ a^2&b^2&c^2\end{pmatrix} = (c-b)(c-a)(b-a)$
And I don't see how they get this. If I calculate the determinant I am always getting this:
$det\begin{pmatrix}1&1&1\\ a&b&c\\ a^2&b^2&c^2\end{pmatrix} = (bc^2-b^2c)-(ac^2-a^2c)+(ab^2-a^2b)=c(b(c-b)-a(c-a)+ab(b-a)).$
But after that point I don't know how to proceed and get the form above. Can you help me?

Subtracting the first colum from the second and third columns we get
$$
\det\begin{pmatrix}1&1&1\\ a& b& c\\a^2& b^2& c^2\end{pmatrix}=\det\begin{pmatrix}1&0&0\\a&b-a&c-a\\ a^2&b^2-a^2&c^2-a^2\end{pmatrix}
$$
It follows that
\begin{eqnarray}
\det\begin{pmatrix}1&1&1\\ a& b& c\\a^2& b^2& c^2\end{pmatrix}&=&\det\begin{pmatrix}1&0&0\\a&b-a&c-a\\ a^2&b^2-a^2&c^2-a^2\end{pmatrix}=\det\begin{pmatrix}b-a&c-a\\ b^2-a^2&c^2-a^2\end{pmatrix}\\
&=&\det\begin{pmatrix}b-a&c-a\\ (b-a)(b+a)&(c-a)(c+a)\end{pmatrix}\\
&=&(b-a)(c-a)\det\begin{pmatrix}1&1\\ b+a&c+a\end{pmatrix}=(b-a)(c-a)(c+a-b-a)\\
&=&(b-a)(c-a)(c-b)
\end{eqnarray}
