Prove the following identity: $\left| {\begin{array}{*{20}{c}}
  a&b&c \\
  {a - b}&{b - c}&{c - a} \\
  {b + c}&{c + a}&{a + b}
\end{array}} \right| = {a^3} + {b^3} + {c^3} - 3abc$

Hint: To prove the given identity, solve the determinant given in the left-side and then compare it to right-side. Use the definition of determinant to expand the left-side. Use parentheses and square brackets to avoid complications and solve by opening parenthesis step by step. Pair the same terms with opposite signs and equate them to zero.

Complete step-by-step answer:
In the given problem, we have to prove the equality between two expressions. On left-side, we have a determinant of three-cross-three matrix, i.e. matrix with three rows and three columns. And on right-side, we have a cubic expression in terms of $a,b{\text{ and }}c$ .
Before starting with the solution, we should understand the concept of determinants and how to evaluate them. In linear algebra, the determinant is a scalar value that can be computed from the elements of a square matrix and encodes certain properties of the linear transformation described by the matrix. The determinant of a matrix A is denoted $\det \left( A \right)$ , $\det A$ , or \[\left| A \right|\] . Geometrically, it can be viewed as the volume scaling factor of the linear transformation described by the matrix.
In the case of a 2 × 2 matrix, the determinant may be defined as:
$\left| A \right| = \left| {\begin{array}{*{20}{c}}
  a&b \\
  c&d
\end{array}} \right| = ad - bc$
Similarly, for a 3 × 3 matrix A, its determinant is:
$\left| A \right| = \left| {\begin{array}{*{20}{c}}
  a&b&c \\
  d&e&f \\
  g&h&i
\end{array}} \right| = a\left| {\begin{array}{*{20}{c}}
  e&f \\
  h&i
\end{array}} \right| - b\left| {\begin{array}{*{20}{c}}
  d&f \\
  g&i
\end{array}} \right| + c\left| {\begin{array}{*{20}{c}}
  d&e \\
  g&h
\end{array}} \right| = aei + bfg + cdh - ceg - bdi - afh$
And for our case, we have a $3 \times 3$ determinant as: $\left| {\begin{array}{*{20}{c}}
  a&b&c \\
  {a - b}&{b - c}&{c - a} \\
  {b + c}&{c + a}&{a + b}
\end{array}} \right|$
So this determinant can also be evaluated similarly by using the above definition:
$ \Rightarrow \left| {\begin{array}{*{20}{c}}
  a&b&c \\
  {a - b}&{b - c}&{c - a} \\
  {b + c}&{c + a}&{a + b}
\end{array}} \right| = a\left| {\begin{array}{*{20}{c}}
  {b - c}&{c - a} \\
  {c + a}&{a + b}
\end{array}} \right| - b\left| {\begin{array}{*{20}{c}}
  {a - b}&{c - a} \\
  {b + c}&{a + b}
\end{array}} \right| + c\left| {\begin{array}{*{20}{c}}
  {a - b}&{b - c} \\
  {b + c}&{c + a}
\end{array}} \right|$
Now, this can be further simplified using the definition of $2 \times 2$ matrix:
$ \Rightarrow a\left| {\begin{array}{*{20}{c}}
  {b - c}&{c - a} \\
  {c + a}&{a + b}
\end{array}} \right| - b\left| {\begin{array}{*{20}{c}}
  {a - b}&{c - a} \\
  {b + c}&{a + b}
\end{array}} \right| + c\left| {\begin{array}{*{20}{c}}
  {a - b}&{b - c} \\
  {b + c}&{c + a}
\end{array}} \right| = $
$ = a\left[ {\left( {b - c} \right)\left( {a + b} \right) - \left( {c + a} \right)\left( {c - a} \right)} \right] - b\left[ {\left( {a - b} \right)\left( {a + b} \right) - \left( {c - a} \right)\left( {b + c} \right)} \right] + c\left[ {\left( {a - b} \right)\left( {c + a} \right) - \left( {b - c} \right)\left( {b + c} \right)} \right]$
Now let’s multiply the terms in parenthesis while being careful with the brackets.
$ = a\left[ {\left( {ba + {b^2} - ca - cb} \right) - \left( {{c^2} - {a^2}} \right)} \right] - b\left[ {\left( {{a^2} - {b^2}} \right) - \left( {cb + {c^2} - ab - ac} \right)} \right] + c\left[ {\left( {ac + {a^2} - bc - ba} \right) - \left( {{b^2} - {c^2}} \right)} \right]$
We can now open these parentheses and make quadratic expressions inside square brackets
$ = a\left[ {{a^2} + {b^2} - {c^2} + ba - ca - cb} \right] - b\left[ {{a^2} - {b^2} - {c^2} - cb + ab + ac} \right] + c\left[ {{a^2} - {b^2} + {c^2} + ac - bc - ba} \right]$
On opening the square brackets, we get:
\[ \Rightarrow {a^3} + a{b^2} - a{c^2} + b{a^2} - c{a^2} - abc - {a^2}b + {b^3} + {c^2}b + c{b^2} - a{b^2} - abc + {a^2}c - {b^2}c + {c^3} + a{c^2} - b{c^2} - abc\]
We can now pair up the same terms with opposite signs and equate them to zero
\[ \Rightarrow {a^3} + {b^3} + {c^3} - abc - abc - abc + \]
\[ + \left( {a{b^2} - a{b^2}} \right) + \left( {a{c^2} - a{c^2}} \right) + \left( {{a^2}b - {a^2}b} \right) + \left( {c{a^2} - c{a^2}} \right) + \left( {{c^2}b - b{c^2}} \right) + \left( {c{b^2} - {b^2}c} \right)\]
Clearly, each of these three pair is equal to zero; therefore we get the value for determinant:
$ \Rightarrow \left| {\begin{array}{*{20}{c}}
  a&b&c \\
  {a - b}&{b - c}&{c - a} \\
  {b + c}&{c + a}&{a + b}
\end{array}} \right| = {a^3} + {b^3} + {c^3} - 3abc$
This is equal to the RHS of the given equation in the problem.
Hence, we proved that: $\left| {\begin{array}{*{20}{c}}
  a&b&c \\
  {a - b}&{b - c}&{c - a} \\
  {b + c}&{c + a}&{a + b}
\end{array}} \right| = {a^3} + {b^3} + {c^3} - 3abc$

Note: The determinant is used for solving linear equations. The determinant can be viewed as a function whose input is a square matrix and whose output is a number.
The use of parentheses and square brackets was the crucial part of the solution. Go step by step and be careful while opening and evaluating brackets. Remember to use $\left( {a + b} \right)\left( {c + d} \right) = ac + ad + bc + bd$ and identity $\left( {a - b} \right)\left( {a + b} \right) = {a^2} - {b^2}$ while simplifying the value of the determinant.