| = - (a3+b3+c3 - 3abc) = - (a + b + c) (a2+b2+c2 - ab - bc - ca) Now a2+b2+c2 - ab - bc - ca ≠0 ∴ the given system of equations can have non-zero solution (x, y, z are not all zero) If D = 0 , i.e. If a + b + c = 0 ∴ Statement 1 is false Again a2+b2+c2 - ab - bc - ca =
1
2
[2a2+2b2+2c2 - 2ab - 2bc - 2ca] =
1
2
[(a−b)2+(b−c)2 + (c−a)2] > 0 [∵ a, b, c are all different] ∴ a2+b2+c2 > ab + bc + ca ∴ Statement is true ∴ D holds