This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.
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Here are the solutions for both problems using properties of determinants. Problem 7: Show that 1 & a & bc \\ 1 & b & ca \\ 1 & c & ab = (a-b)(b-c)(c-a). Step 1: Apply row operations to simplify the determinant. Let R_1, R_2, R_3 be the first, second, and third rows. Apply the operations R_2 R_2 - R_1 and R_3 R_3 - R_1. 1 & a & bc \\ 1 & b & ca \\ 1 & c & ab = 1 & a & bc \\ 1-1 & b-a & ca-bc \\ 1-1 & c-a & ab-bc = 1 & a & bc \\ 0 & b-a & c(a-b) \\ 0 & c-a & b(a-c) Step 2: Expand the determinant along the first column. = 1 · ( (b-a) · b(a-c) - c(a-b) · (c-a) ) = (b-a)b(a-c) - c(a-b)(c-a) Step 3: Factor and simplify the expression. Note that b-a = -(a-b) and a-c = -(c-a). Substitute these into the expression: = (-(a-b)) · b · (-(c-a)) - c · (a-b) · (c-a) = (a-b)b(c-a) - c(a-b)(c-a) Factor out the common term (a-b)(c-a): = (a-b)(c-a)(b-c) Rearrange the terms to match the target expression: = (a-b)(b-c)(c-a) Thus, we have shown that: 1 & a & bc \\ 1 & b & ca \\ 1 & c & ab = (a-b)(b-c)(c-a) Problem 8: Show that 1+a & 1 & 1 \\ 1 & 1+b & 1 \\ 1 & 1 & 1+c = abc+ab+bc+ca. Step 1: Apply row operations to simplify the determinant. Let R_1, R_2, R_3 be the first, second, and third rows. Apply the operations R_2 R_2 - R_1 and R_3 R_3 - R_1. 1+a & 1 & 1 \\ 1 & 1+b & 1 \\ 1 & 1 & 1+c = 1+a & 1 & 1 \\ 1-(1+a) & (1+b)-1 & 1-1 \\ 1-(1+a) & 1-1 & (1