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 ofare \[-1,\,1-2\omega \] and\[1-2{{\omega }^{2}}\] Note If 1,andare the cube roots of unity, then \[1+\omega +{{\omega }^{2}}=0\] and \[{{\omega }^{3}}=1\].You need to login to perform this action.
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