Evaluate the following determinant
$\left| {\begin{array}{*{20}{c}}
  1&a&{bc} \\
  1&b&{ca} \\
  1&c&{ab}
\end{array}} \right|$

Hint: For evaluating the given determinant, we can apply row or column operations to make the expansion easier.

Complete step-by-step answer:
$ \Rightarrow $ Let $\vartriangle = $ $\left| {\begin{array}{*{20}{c}}
  1&a&{bc} \\
  1&b&{ca} \\
  1&c&{ab}
\end{array}} \right|$
Now, here we will apply row operations for solving the above determinant.
Now, applying row operation in ${R_2}$
$ \Rightarrow {R_2} \to {R_2} - {R_1}$
\[ \Rightarrow \vartriangle = \left| {\begin{array}{*{20}{c}}
  1&a&{bc} \\
  0&{b - a}&{ca - bc} \\
  1&c&{ab}
\end{array}} \right|\]
Now, applying row operation in R₃
$ \Rightarrow {R_3} \to {R_3} - {R_1}$
\[ \Rightarrow \vartriangle = \left| {\begin{array}{*{20}{c}}
  1&a&{bc} \\
  0&{b - a}&{ca - bc} \\
  0&{c - a}&{ab - bc}
\end{array}} \right| = \left| {\begin{array}{*{20}{c}}
  1&a&{bc} \\
  0&{b - a}&{c\left( {a - b} \right)} \\
  0&{c - a}&{b\left( {a - c} \right)}
\end{array}} \right|\]
Now, for making the expansion easy we can take common terms out from the determinant. So taking (a – b) and (a – c) common from ${R_2}$ and ${R_3}$ respectively, we get,
\[ \Rightarrow \vartriangle = \left( {a - b} \right)\left( {a - c} \right)\left| {\begin{array}{*{20}{c}}
  1&a&{bc} \\
  0&{ - 1}&c \\
  0&{ - 1}&b
\end{array}} \right|\]
Now we can expand the determinant along any row or column.
So, here it is better to expand the above determinant along the first column as two elements of the first column are 0. So, it will make our calculations easier.
So, expanding along first column, we get,
\[ \Rightarrow \vartriangle = \left( {a - b} \right)\left( {a - c} \right)\left| {\begin{array}{*{20}{c}}
  { - 1}&c \\
  { - 1}&b
\end{array}} \right| = \left( {a - b} \right)\left( {a - c} \right)\left( { - b + c} \right)\]
$ \Rightarrow $ Hence $\left| {\begin{array}{*{20}{c}}
  1&a&{bc} \\
  1&b&{ca} \\
  1&c&{ab}
\end{array}} \right| = \vartriangle = \left( {a - b} \right)\left( {a - c} \right)\left( {c - b} \right)$

Note: Whenever we come up with these types of problems, to make calculations easy, first reduce the determinant by applying row or column operations then expand it. Take care of the signs while carrying out expansions. It is better to expand along a row or column with most 0’s.