question_answer

If the cube roots of unity are\[1,\omega ,{{\omega }^{2}}\]then the roots of the equation \[{{(x-1)}^{3}}+8=0,\] are     AIEEE  Solved  Paper-2005

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 \[{{\omega }^{3}}=1\].