If a, b, c are real numbers, then prove that |(a,b,c)(b,c,a)(c,a,b)| = - (a + b + c)(a + bω + cω^2)(a + bω^2 + cω)

Question

If a, b, c are real numbers, then prove that 

\(\begin{vmatrix} a & b &c \\[0.3em] b & c & a \\[0.3em] c & a &b \end{vmatrix}\)= - (a + b + c)(a + bω + cω²)(a + bω² + cω) 

Where ω is a complex number and cube root of unity.

Answer 1

Let \(\Delta=\) \(\begin{vmatrix} a & b &c \\[0.3em] b & c & a \\[0.3em] c & a &b \end{vmatrix}\) 

= (a + b + c) {- (b - c)² - (a - c) (a - b)}

LHS = - (a + b + c) (a² + b² + c² - ab - bc - ca)

Also, RHS,