If a- b-c are positive integers and - is imaginary cube root of unity and fx-x6a-x6b-1-x6c-2- then f- equals-

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Standard XII

Mathematics

Complex Numbers

If a, b,c are...

Question

If a, b,c are positive integers and $ω$ is imaginary cube root of unity and $f (x) = x^{6 a} + x^{6 b + 1} + x^{6 c + 2}$ , then $f (ω)$ equals:

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Solution

The correct option is A $0$
$f (ω) = ω^{6 a} + ω^{6 b + 1} + ω^{6 c + 2}$
$= (ω^{6})^{a} + (ω^{6})^{b} . ω + (ω^{6})^{c} . ω^{2}$
$= 1 + ω + ω^{2} [∵ ω^{6} = (ω^{3})^{2} = 1]$
$= 0$ $[∵ 1 + ω + ω^{2} = 0]$

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If a, b,c are positive integers and ω is imaginary cube root of unity and f(x)=x6a+x6b+1+x6c+2, then f(ω) equals:

The correct option is A 0f(ω)=ω6a+ω6b+1+ω6c+2=(ω6)a+(ω6)b.ω+(ω6)c.ω2=1+ω+ω2[∵ω6=(ω3)2=1]=0 [∵1+ω+ω2=0]

If a, b,c are positive integers and $ω$ is imaginary cube root of unity and $f (x) = x^{6 a} + x^{6 b + 1} + x^{6 c + 2}$ , then $f (ω)$ equals:

The correct option is A $0$
$f (ω) = ω^{6 a} + ω^{6 b + 1} + ω^{6 c + 2}$
$= (ω^{6})^{a} + (ω^{6})^{b} . ω + (ω^{6})^{c} . ω^{2}$
$= 1 + ω + ω^{2} [∵ ω^{6} = (ω^{3})^{2} = 1]$
$= 0$ $[∵ 1 + ω + ω^{2} = 0]$