Using properties of determinants, prove the following: \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)

LHS=\begin{vmatrix} 1 & 1 & 1 \ a & b & c \ a^3 & b^3 & c^3 \end{vmatrix}
Applying C_2 \rightarrow C_2 - C_1, C_3 \rightarrow C_3 - C_1, we get
=\begin{vmatrix} 1 & 0 & 0 \ a & b-a & c-a \ a^3 & b^3-a^3 & c^3-a^3 \end{vmatrix}
Taking out (b - a), (c - a) common from C_2 and C_3 respectively, we get
=(b-a)(c-a)\begin{vmatrix} 1 & 0 & 0 \ a & 1 & 1 \ a^3 & b^2+ab+a^2 & c^2+ac+a^2 \end{vmatrix}
Expanding along R_1, we get
=(b-a)(c-a)[1(c^2+ac+a^2 - b^2-ab-a^2) - 0 + 0]
=(b-a)(c-a)(c^2+ac-b^2-ab)
=(b-a)(c-a)[-(b^2-c^2)-a(b-c)]
=(b-a)(c-a)[(b-c)(-b-c-a)]
=(a-b)(b-c)(c-a)(a+b+c)