• question_answer If the cube roots of unity be $1,\omega ,{{\omega }^{2}},$ then the roots of the equation ${{(x-1)}^{3}}+8=0$are [IIT 1979; MNR 1986; DCE 2000; AIEEE 2005] A) $-1,\,1+2\omega ,\,1+2{{\omega }^{2}}$ B) $-1,\,1-2\omega ,\,1-2{{\omega }^{2}}$ C) $-1,\,-1,\,-1$ D) None of these

${{(x-1)}^{3}}=-8\Rightarrow x-1={{(-8)}^{1/3}}$ Þ $x-1=-2,-2\omega ,-2{{\omega }^{2}}$ Þ $x=-1,1-2\omega ,1-2{{\omega }^{2}}$ Trick: By inspection, we see that (b) satisfies the equation i.e,   ${{(-1-1)}^{3}}+8=0,{{(1-2\omega -1)}^{3}}+8=0$${{(1-2{{\omega }^{2}}-1)}^{3}}+8=0$.