Word Count : 518

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To prove that the determinant \[ \begin{vmatrix} 1 & bc & bc(b+c) \\ 1 & ca & ca(c+a) \\ 1 & ab & ab(a+b) \end{vmatrix} \] is independent of \( a, b, c \), we will compute it step by step. --- ### Step 1: Expand the determinant We expand the determinant along the first column: \[ \begin{vmatrix} 1 & bc & bc(b+c) \\ 1 & ca & ca(c+a) \\ 1 & ab & ab(a+b) \end{vmatrix} = 1 \cdot \begin{vmatrix} ca & ca(c+a) \\ ab & ab(a+b) \end{vmatrix} - 1 \cdot \begin{vmatrix} bc & bc(b+c) \\ ab & ab(a+b) \end{vmatrix} + 1 \cdot \begin{vmatrix} bc & bc(b+c) \\ ca & ca(c+a) \end{vmatrix}. \] --- ### Step 2: Compute the \( 2 \times 2 \) determinants Now compute each \( _____ ___ ____ _____ _________ ______ ___ _______ ___ ________ ________ __________.
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