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GATE CSE 1995 | Question: 2.8

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If the cube roots of unity are $1, \omega$ and $\omega^2$, then the roots of the following equation are $$(x-1)^3 +8 =0$$

  1. $-1, 1 + 2\omega, 1 + 2\omega^2$   
  2. $1, 1 - 2\omega, 1 - 2\omega^2$
  3. $-1, 1 - 2\omega, 1 - 2\omega^2$
  4. $-1, 1 + 2\omega, -1 + 2\omega^2$    
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3 Answers

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The given equation is: $(x-1)^{3}=-8$

or,                                   $((x-1)/-2)^{3}=1$

Let  $((x-1)/-2)= Z$

So, the equation will be changed to: $Z^{3}=1$
The roots will be $Z=1, ω, ω^{2}$

Putting back the value of $Z$

$\quad (x-1)/-2 = 1, ω, ω^{2}$
$\quad \implies x= -1, 1-2$ω$, 1-2ω^{2}$

Correct Answer: $C$
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Answer is C,
Just put values of C in place of $x.$ It will satisfy the equation.
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Putting the values provided in the options and keeping in mind that w^3=1 and 1+w+w^2=0 we find option C is correct.
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