Show that 1 & a & bc \ 1 & b & ca \ 1 & c & ab = (a−b)(b−c)(c−a).

Here are the solutions for both problems using properties of determinants.

:
Show that $\begin{vmatrix} 1 & a & bc \\ 1 & b & ca \\ 1 & c & ab \end{vmatrix} = (a-b)(b-c)(c-a)$.

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 $R_2 \to R_2 - R_1$ and $R_3 \to R_3 - R_1$.

$$\begin{vmatrix} 1 & a & bc \\ 1 & b & ca \\ 1 & c & ab \end{vmatrix} = \begin{vmatrix} 1 & a & bc \\ 1-1 & b-a & ca-bc \\ 1-1 & c-a & ab-bc \end{vmatrix}$$ $$= \begin{vmatrix} 1 & a & bc \\ 0 & b-a & c(a-b) \\ 0 & c-a & b(a-c) \end{vmatrix}$$

Step 2: Expand the determinant along the first column.

$$= 1 \cdot \left( (b-a) \cdot b(a-c) - c(a-b) \cdot (c-a) \right)$$ $$= (b-a)b(a-c) - c(a-b)(c-a)$$

Step 3: Factor and simplify the expression.
Note that $b-a = -(a-b)$ and $a-c = -(c-a)$.
Substitute these into the expression:

$$= (-(a-b)) \cdot b \cdot (-(c-a)) - c \cdot (a-b) \cdot (c-a)$$ $$= (a-b)b(c-a) - c(a-b)(c-a)$$

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:

$$\boxed{\begin{vmatrix} 1 & a & bc \\ 1 & b & ca \\ 1 & c & ab \end{vmatrix} = (a-b)(b-c)(c-a)}$$

:
Show that $\begin{vmatrix} 1+a & 1 & 1 \\ 1 & 1+b & 1 \\ 1 & 1 & 1+c \end{vmatrix} = abc+ab+bc+ca$.

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 $R_2 \to R_2 - R_1$ and $R_3 \to R_3 - R_1$.

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