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Using the properties of determinants, prove that:

1 bc bc(b+c)

1 ca ca(c+a) = 0

1 ab ab(a+b)

$| \begin{matrix} 1 & b c & b c (b + c) \\ 1 & c a & c a (c + a) \\ 1 & a b & a b (a + b) \end{matrix} | R_{2} \to R_{2} - R_{1} a n d R_{3} \to R_{3} - R_{1} = | \begin{matrix} 1 & b c & b c (b + c) \\ 0 & c (a - b) & c^{2} (a - b) + c (a - b) (a + b) \\ 0 & b (a - c) & b^{2} (a - c) + b (a - c) (a + c) \end{matrix} | = (a - b) (a - c) | \begin{matrix} 1 & b c & b c (b + c) \\ 0 & c & c^{2} + a c + b c \\ 0 & b & b^{2} + a b + b c \end{matrix} | = (a - b) (a - c) [b^{2} c + a b c + b c^{2} - b c^{2} - a b c - b^{2} c] = 0$

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