Prove that: \[\left| {\begin{array}{*{20}{c}}
  {{a^2} + 1}&{ab}&{ac} \\
  {ab}&{{b^2} + 1}&{bc} \\
  {ac}&{bc}&{{c^2} + 1}
\end{array}} \right| = {a^2} + {b^2} + {c^2} + 1\]

Hint: Here we simply have to find determinant of a 3 by 3 matrix.
\[\left| {\begin{array}{*{20}{c}}
  {{a_{11}}}&{{a_{22}}}&{{a_{33}}} \\
  {{b_{11}}}&{{b_{22}}}&{{b_{33}}} \\
  {{c_{11}}}&{{c_{22}}}&{{c_{33}}}
\end{array}} \right| = {a_{11}}\left| {\begin{array}{*{20}{c}}
  {{b_{22}}}&{{b_{33}}} \\
  {{c_{22}}}&{{c_{33}}}
\end{array}} \right| - {a_{22}}\left| {\begin{array}{*{20}{c}}
  {{b_{11}}}&{{b_{33}}} \\
  {{c_{11}}}&{{c_{33}}}
\end{array}} \right| + {a_{33}}\left| {\begin{array}{*{20}{c}}
  {{b_{11}}}&{{b_{22}}} \\
  {{c_{11}}}&{{c_{22}}}
\end{array}} \right|\]

Complete step-by-step answer:
Given that,
\[\left| {\begin{array}{*{20}{c}}
  {{a^2} + 1}&{ab}&{ac} \\
  {ab}&{{b^2} + 1}&{bc} \\
  {ac}&{bc}&{{c^2} + 1}
\end{array}} \right|\]
Now in order to determine the determinant value we will start with \[{a_{11}}\] (\[{a^2} + 1\]) element, and we will drop first row and first column ,then \[{a_{22}}\] (\[ab\]) element and drop second row ,second column and at last take \[{a_{33}}\] (\[ac\]) element and drop third row and third column, as shown below.
\[ \Rightarrow ({a^2} + 1)\left| {\begin{array}{*{20}{c}}
  {{b^2} + 1}&{bc} \\
  {bc}&{{c^2} + 1}
\end{array}} \right| - ab\left| {\begin{array}{*{20}{c}}
  {ab}&{bc} \\
  {ac}&{{c^2} + 1}
\end{array}} \right| + ac\left| {\begin{array}{*{20}{c}}
  {ab}&{{b^2} + 1} \\
  {ac}&{bc}
\end{array}} \right|\]
Now solve the inner determinant as
\[\left| {\begin{array}{*{20}{c}}
  a&c \\
  b&d
\end{array}} \right| = ad - bc\]
Applying this to above determinant
\[ \Rightarrow ({a^2} + 1)\left( {\left( {{b^2} + 1} \right)\left( {{c^2} + 1} \right) - \left( {bc} \right)\left( {bc} \right)} \right) - ab\left( {ab\left( {{c^2} + 1} \right) - acbc} \right) + ac\left( {abbc - ac\left( {{b^2} + 1} \right)} \right)\]
Multiplying the term sin the brackets,
\[ \Rightarrow ({a^2} + 1)\left( {{b^2}{c^2} + {b^2} + {c^2} + 1 - {b^2}{c^2}} \right) - ab\left( {ab{c^2} + ab - ab{c^2}} \right) + ac\left( {a{b^2}c - ac{b^2} - ac} \right)\]
Now if we observe the terms in red colour obtained after solving the brackets can be cancelled
\[ \Rightarrow ({a^2} + 1)\left( {{b^2} + {c^2} + 1} \right) - ab\left( {ab} \right) + ac\left( { - ac} \right)\]
Now multiply the terms outside the brackets with the terms inside respectively.
\[ \Rightarrow {a^2}{b^2} + {a^2}{c^2} + {a^2} + {b^2} + {c^2} + 1 - {a^2}{b^2} - {a^2}{c^2}\]
Again after this operation if we observe the terms in red colour can be cancelled.
\[ \Rightarrow {a^2} + {b^2} + {c^2} + 1\]
This is our final answer and it is equal to RHS.
Thus we proved the statement.

Note: Note that when we solve a determinant the order of elements is very important. If we miss a single element it will lead to a wrong answer. Always perform each and every step of this type of problem