If ω is complex cube root of unity and a, b, c are three real numbers such that 1a+ω+1b+ω+1c+ω=2ω2 and 1a+ω2+1b+ω2+1c+ω2=2ω , then1a+1+1b+1+1c+1=

( Since ω2=1ω   and  ω=1ω2  ) the given relation may be rewritten as 1a+ω+1b+ω+1c+ω=2ω and  1a+ω2+1b+ω2+1c+ω2=2ω2 Clearly ω  and ω2  are the roots of 1a+x+1b+x+1c+x=2x  or (b+x)(c+x)+(a+x)(c+x)+(a+x)(b+x)(a+x)(b+x)(c+x)=2x or  x[3x2+2(a+b+c)x+bc+ca+ab] =2[abc+(bc+ca+ab)x(a+b+c)x2+x3] ⇒x3−(bc+ca+ab)x−2abc=0 Now if α is the third root of this equation then sum of the roots,α+ω+ω2=0⇒α=1 Hence,1 is the root of equation(1) we get 1a+1+1b+1+1c+1=2