Let $\Delta=\left|\begin{array}{lll}1 & 1 & 1 \\ a & b & c \\ a^{3} & b^{3} & c^{3}\end{array}\right|$

Applying $C_{1} \rightarrow C_{1}-C_{3}$ and $C_{2} \rightarrow C_{2}-C_{3},$ we have:

$\Delta=\left|\begin{array}{ccc}0 & 0 & 1 \\ a-c & b-c & c \\ a^{3}-c^{3} & b^{3}-c^{3} & c^{3}\end{array}\right|$

$=\left|\begin{array}{ccc}0 & 0 & 1 \\ a-c & b-c & c \\ (a-c)\left(a^{2}+a c+c^{2}\right) & (b-c)\left(b^{2}+b c+^{2}\right) & c^{3}\end{array}\right|$

$ = (c – a)(b – c)\left| {\begin{array}{*{20}{c}}
  0&0&1 \\ 
  { – 1}&1&c \\ 
  { – ({a^2} + ac + {c^2})}&{({b^2} + bc + {c^2})}&{{c^3}} 
\end{array}} \right|$

Applying $C_{1} \rightarrow C_{1}+C_{2},$ we have:

$\Delta=(c-a)(b-c)\left|\begin{array}{ccc}0 & 0 & 1 \\ 0 & 1 & c \\ \left(b^{2}-a^{2}\right)+(b c-a c) & \left(b^{2}+b c+c^{2}\right) & c^{3}\end{array}\right|$

$=(b-c)(c-a)(a-b)\left|\begin{array}{ccc}0 & 0 & 1 \\ 0 & 0 & c \\ -(a+b+c) & \left(b^{2}+b c+c^{2}\right) & c^{3}\end{array}\right|$

$=(a-b)(b-c)(c-a)(a+b+c)\left|\begin{array}{ccc}0 & 0 & 1 \\ 0 & 1 & c \\ -1 & \left(b^{2}+b c+c^{2}\right) & c^{3}\end{array}\right|$

Expanding along $C_{1}$, we have:

$\Delta=(a-b)(b-c)(c-a)(a+b+c)(-1)\left|\begin{array}{cc}0 & 1 \\ 1 & c\end{array}\right|$

$=(a-b)(b-c)(c-a)(a+b+c)$

Hence, the given result is proved.