    Question

# If ω(≠1) is cube root of unity satisfying 1a+ω+1b+ω+1c+ω=2ω2 and 1a+ω2+1b+ω2+1c+ω2=2ω, then the value of 1a+1+1b+1+1c+1 is :

A
2
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B
ω
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C
1
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D
2
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Solution

## The correct option is A 2We have ω(≠1) is cube root of unity satisfying 1a+ω+1b+ω+1c+ω=2ω2=2ω and 1a+ω2+1b+ω2+1c+ω2=2ω=2ω2 ∴ ω and ω2 are roots of the equation 1a+x+1b+x+1c+x=2x⋯(1) Simplifying the equation, we get ⇒x[(x+a)(x+c)+(x+a)(x+b)+(x+b)(x+c)] =2(x+a)(x+b)(x+c)⇒x[3x2+2(a+b+c)x+ab+bc+ca] =2[x3+(a+b+c)x2+(ab+bc+ca)x+abc]⇒x3−(ab+bc+ca)x−2abc=0 If the roots of the cubic equations are ω,ω2,γ, so from sum of the roots ω+ω2+γ=0⇒γ=1 Hence putting x=1 in (1), we get 1a+1+1b+1+1c+1=2  Suggest Corrections  0      Similar questions
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