A) \[-1,1+2\omega ,1+2{{\omega }^{2}}\]
B) \[-1,1-2\omega ,1-2{{\omega }^{2}}\]
C) \[-1,-1,-1\]
D) \[-1,-1+2\omega ,-1-2{{\omega }^{2}}\]
Correct Answer: B
Solution :
Since, \[{{(x-1)}^{3}}+8=0\] \[\Rightarrow \]\[{{(x-1)}^{3}}=-8={{(-2)}^{3}}\] \[\Rightarrow \] \[{{\left( \frac{x-1}{-2} \right)}^{3}}=1\] \[\Rightarrow \] \[\left( \frac{x-1}{-2} \right)={{(1)}^{1/3}}\] \[\therefore \]Cube roots of\[\left( \frac{x-1}{-2} \right)\]are 1,\[\omega \] and \[{{\omega }^{2}}\]. \[\Rightarrow \]Cube roots of \[(x-1)\] are \[-2,\,-2\,\omega \] and\[-2{{\omega }^{2}}\]. \[\Rightarrow \]Cube roots ofYou need to login to perform this action.
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