$$\begin{vmatrix}
1 & 1 & 1\\
a & b & c\\
a^3 & b^3 & c^3
\end{vmatrix} =
(a-b)(b-c)(c-a)(a+b+c)$$
we have to solve this by using the properties of determinants without actually expanding the determinant. I am Unable to think which calculation to apply so was hoping for an hint.
Best Answer
Using $C_2'=C_2-C_1, C_3'=C_3-C_1$
$$\begin{vmatrix} 1 & 1 & 1\\ a & b & c\\ a^3 & b^3 & c^3 \end{vmatrix}$$
$$=\begin{vmatrix} 1 & 0 & 0\\ a & b-a & c-a\\ a^3 & b^3-a^3 & c^3-a^3 \end{vmatrix}$$
$$=-(a-b)(c-a)\begin{vmatrix} 1 & 0 & 0\\ a & 1 & 1\\ a^3 & b^2+ab+a^2 & c^2+ca+a^2 \end{vmatrix}$$
Use $C_2'=C_2-C_3$