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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. $-1, 1 + 2\omega, -1 + 2\omega^2$    
asked in Set Theory & Algebra by Veteran (59.7k points)
edited by | 591 views

2 Answers

+16 votes
Best answer
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$, 1-2ω^{2}$
answered by Active (3.7k points)
selected by
0
nice approach srivivek95
0
good answer vivek
+14 votes
Answer is C,
Just put values of C in place of $x.$ iIt will satisfy the equation.
answered by Loyal (8.1k points)
+11

one must know

  • 1+ω+ω2 =0
  • ω3 = 1

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