Using properties of the determinant, prove the following

\[\left| {\begin{array}{*{20}{c}}
  1&1&1 \\
  a&b&c \\
  {{a^3}}&{{b^3}}&{{c^3}}
\end{array}} \right| = \left( {a - b} \right)\left( {b - c} \right)\left( {c - a} \right)\left( {a + b + c} \right)\]

Hint: Use the necessary row and column transformation in such a way that we can make some common elements of the right hand side. Using the column transformation as${C_2} \to {C_2} - {C_1}$ ,we will be getting the common terms as $\left( {b - c} \right)\left( {c - a} \right)$. Then by using identities and elementary operations in row and column, we can make the LHS side equal to RHS side.

Complete step by step answer:
Now, the left hand side of the given question is given as
\[\left| {\begin{array}{*{20}{c}}
  1&1&1 \\
  a&b&c \\
  {{a^3}}&{{b^3}}&{{c^3}}
\end{array}} \right|\]
Now we will apply the column transformation as ${C_2} \to {C_2} - {C_1}$ ,
${C_3} \to {C_3} - {C_1}$ , we get the determinant or the LHS as
$LHS = \left| {\begin{array}{*{20}{c}}
  1&0&0 \\
  a&{\left( {b - a} \right)}&{\left( {c - a} \right)} \\
  {{a^3}}&{{b^3} - {a^3}}&{{c^3} - {a^3}}
\end{array}} \right|$
Taking out the common terms from column number 2 and 3 respectively as $\left( b-a \right)\left( c-a \right)$ , the equation in the left hand side becomes
$\left( b-a \right)\left( c-a \right)$ $\left| \begin{matrix}
   1 & 0 & 0 \\
   a & 1 & 1 \\
   {{a}^{3}} & {{b}^{2}}+ab+{{a}^{2}} & {{c}^{2}}+ac+{{a}^{2}} \\
\end{matrix} \right|$
Now further expanding along the row number 1, the equation becomes
$LHS=-\left( a-b \right)\left( c-a \right)\left[ 1\left( {{c}^{2}}+ac+{{a}^{2}}-{{b}^{2}}-ab-{{a}^{{{2}^{{}}}}} \right) \right]$
Again, on further simplification we get, the equation pertaining to the left hand side as
$=-\left( a-b \right)\left( c-a \right)\left( {{c}^{2}}+ac-{{b}^{2}}-ab \right)$
Again, on further simplifying, we get
$=-\left( a-b \right)\left( c-a \right)\left( -\left( {{b}^{2}}-{{c}^{2}} \right)-a\left( b-c \right) \right)$
On further expansion
$-\left( a-b \right)\left( c-a \right)\left( b-c \right)\left( -b-c-a \right)$
Now the LHS side becomes
$\left( a-b \right)\left( b-c \right)\left( c-a \right)\left( a+b+c \right)$
Now the left hand side becomes equal to the right hand side.

Note:
The elementary transformation should be done in such a way that we get some common elements from the RHS side, here in this case the common element is $\left( b-c \right)\left( c-a \right)$. Be careful while doing the mathematical operations in the elementary transformation as the student can usually commit a mistake in performing them. One silly mistake in the elementary transformation can create confusion in the mind and the student will leave the solution in between as the left hand side could not be verified with the right hand side.