If a-b-c are roots of y3 -3y2-3y-26- 0 and - is cube roots of unity- then the value of a-1-b-1-b-1-c-1 -c-1-a-1 equals

$\frac{1}{a + ω}$ + $\frac{1}{b + ω}$ + $\frac{1}{c + ω}$ = 2 $ω^{2}$ and $\frac{1}{a + ω^{2}}$ + $\frac{1}{b + ω^{2}}$ + $\frac{1}{c + ω^{2}}$ = 2 $ω$ , then $\frac{1}{a + 1}$ + $\frac{1}{b + 1}$ + $\frac{1}{c + 1}$ is equal to:

Q. If

ω

be complex cube root of unity satisfying the equation

\frac{1}{a + ω} + \frac{1}{b + ω} + \frac{1}{c + ω} = 2 ω^{2}

and

\frac{1}{a + ω^{2}} + \frac{1}{b + ω^{2}} + \frac{1}{c + ω^{2}} + = 2 ω

, then

\frac{1}{a + 1} + \frac{1}{b + 1} + \frac{1}{c + 1}

is equal to

Q. If

1, ω, ω^{2}

are the cube roots of unity then

(3 + 3 ω^{2} + 5 ω)^{6} - (2 + 6 ω^{2} + 2 ω)^{3}

is equal to

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If a,b,c are roots of y3−3y2+3y+26=0 and ω is cube roots of unity, then the value of a−1b−1+b−1c−1+c−1a−1 equals

The correct option is B 3ω2From given equation (y−1)3=−27(y−1−3)3=1∴y−1−3=(1)1/3∴y−1−3=1,ω,ω2 as a,b,c are roots ∴a−1=−3,b−1=−3ω,c−1=−3ω2 ∴a−1b−1+b−1c−1+c−1a−1=1ω+1ωω2ω3=ω2+ω2+ω2=3ω2

If a,b,c are roots of $y^{3} - 3 y^{2} + 3 y + 26 = 0$ and $ω$ is cube roots of unity, then the value of $\frac{a - 1}{b - 1} + \frac{b - 1}{c - 1} + \frac{c - 1}{a - 1}$ equals