If vectors a→,b→ and c→ are non-coplanar and l, m and n are distinct scalars, then [(la→+mb→+nc→)(lb→+mc→+na→)(lc→+ma→+nb→)]=0 implies

V→1=la→+mb→+nc→V→2=na→+lb→+mc→V→3=ma→+nb→+lc→ when a→,b→ and c→ are non-coplanar. Therefore,V→1V→2V→3=lmnnlmmnl=0or (l+m+n)(l−m)2+(m−n)2+(n−l)2=0or  l+m+n=0---iObviously, l2+ mx + n=0 is satisfied by x=1 due to (i). l3+m3+n3=3lmn(l+m+n)l2+m2+n2−lm−mn−ln=0which is true