\[\left| \,\begin{matrix} bc & b{c}'+{b}'c & {b}'{c}' \\ ca & c{a}'+{c}'a & {c}'{a}' \\ ab & a{b}'+{a}'b & {a}'{b}' \\ \end{matrix}\, \right|\] is equal to

\[(a{b}'+{a}'b)(b{c}'+{b}'c)(c{a}'+{c}'a)\]

\[(ab-{a}'{b}')(bc-{b}'{c}')(ca-{c}'{a}')\]

\[(ab+{a}'{b}')(bc+{b}'{c}')(ca+{c}'{a}')\]

\[(a{b}'-{a}'b)(b{c}'-{b}'c)(c{a}'-{c}'a)\]

\[\left| \,\begin{matrix} bc & b{c}'+{b}'c & {b}'{c}' \\ c

1.	\[\left\| \,\begin{matrix} bc & b{c}'+{b}'c & {b}'{c}' \\ ca & c{a}'+{c}'a & {c}'{a}' \\ ab & a{b}'+{a}'b & {a}'{b}' \\ \end{matrix}\, \right\|\] is equal to
A.	\[(ab-{a}'{b}')(bc-{b}'{c}')(ca-{c}'{a}')\]
B.	\[(ab+{a}'{b}')(bc+{b}'{c}')(ca+{c}'{a}')\]
C.	\[(a{b}'-{a}'b)(b{c}'-{b}'c)(c{a}'-{c}'a)\]
D.	\[(a{b}'+{a}'b)(b{c}'+{b}'c)(c{a}'+{c}'a)\]
Answer» D. \[(a{b}'+{a}'b)(b{c}'+{b}'c)(c{a}'+{c}'a)\]

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\[\left| \,\begin{matrix} bc & b{c}'+{b}'c & {b}'{c}' \\ ca & c{a}'+{c}'a & {c}'{a}' \\ ab & a{b}'+{a}'b & {a}'{b}' \\ \end{matrix}\, \right|\] is equal to

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