To solve the problem, we must evaluate the determinant of the 3x3 matrix:
\( A = \begin{vmatrix} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{vmatrix} \)
The determinant of this matrix type is evaluated using the formula:
\( A = 1 \cdot \begin{vmatrix} b & b^2 \\ c & c^2 \end{vmatrix} - a \cdot \begin{vmatrix} 1 & b^2 \\ 1 & c^2 \end{vmatrix} + a^2 \cdot \begin{vmatrix} 1 & b \\ 1 & c \end{vmatrix} \)
Calculate each 2x2 determinant:
\( \begin{vmatrix} b & b^2 \\ c & c^2 \end{vmatrix} = bc^2 - b^2c = c \cdot b(c - b) \)
\( \begin{vmatrix} 1 & b^2 \\ 1 & c^2 \end{vmatrix} = 1\cdot c^2 - 1\cdot b^2 = c^2 - b^2 = (c-b)(c+b) \)
\( \begin{vmatrix} 1 & b \\ 1 & c \end{vmatrix} = 1\cdot c - 1\cdot b = c - b \)
Substitute these back into the equation for \( A \):
\( A = 1 \cdot c \cdot b(c - b) - a \cdot (c-b)(c+b) + a^2 \cdot (c-b) \)
Factor out \((c-b)\) from each term:
\( A = (c-b) \left( bc - ab - a(b+c) + a^2 \right) \)
Simplify the expression within the parentheses:
\( = (c-b) \left( bc - ab - ab - ac + a^2 \right) \)
\( = (c-b) \left( a^2 - a(b + c) + bc \right) \)
The expression can be factored further to yield:
\( A = (a-b)(b-c)(c-a) \)
The final answer is: