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Let real numbers $a,b,c\in\mathbb{R}$ be given. Without expanding the determinants, prove that the determinants of the matrices $$\begin{bmatrix} 1&a&bc\\ 1&b&ac\\ 1&c&ab \end{bmatrix} \mbox{and} \begin{bmatrix} 1&a&a^2\\ 1&b&b^2\\ 1&c&c^2 \end{bmatrix}$$ are equal.

My first attempt to solve this was to row reduce both matrices to upper triangular and show that the products of the diagonals are equal. This, however, depends on $a,b,c$ all being non-zero. My next attempt was to apply row operations to show that one matrix is just an even permutation of the columns of the other matrix, but this method has the same issue. Maybe there's some way to look at divisors of the characteristic polynomials and derive the desired conclusion that way, but I'm not seeing it.

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It turns out that$$\begin{pmatrix}1&a&a^2\\1&b&b^2\\1&c&c^2\end{pmatrix}=\begin{pmatrix}1&a&bc\\1&b&ac\\1&c&ab\end{pmatrix}.\begin{pmatrix}1&0&-ab-ac-bc\\0&1&a+b+c\\0&0&1\end{pmatrix}.$$Since, obviously,$$\begin{vmatrix}1&0&-ab-ac-bc\\0&1&a+b+c\\0&0&1\end{vmatrix}=1,$$the two given matrices have the same determinant.

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Note that $$\begin{vmatrix} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2\\ \end{vmatrix}-\begin{vmatrix} 1 & a & bc \\ 1 & b & ca \\ 1 & c & ab \\ \end{vmatrix}=\begin{vmatrix} 1 & a & a^2-bc \\ 1 & b & b^2-ca \\ 1 & c & c^2-ab \\ \end{vmatrix}.$$ Then, subtracting the first row from the second and third rows, it follows that $$\begin{vmatrix} 1 & a & a^2-bc \\ 1 & b & b^2-ca \\ 1 & c & c^2-ab \\ \end{vmatrix} =\begin{vmatrix} 1 & a & a^2-bc \\ 0 & b-a & (b-a)(b+a+c) \\ 0 & c-a & (c-a)(c+a+b) \\ \end{vmatrix}.$$ ($\textit{Note}$: Elementary row operations can change the determinant, but adding/subtracting multiples of one row/column to another do not change the determinant.) Because the second and third rows are both multiples of $\begin{bmatrix}0&1&a+b+c\end{bmatrix}$, they are linearly dependent. Hence, $0$ is an eigenvalue of this matrix, and thus the determinant is $0$. The desired result follows.

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