Question

lf the cube roots of unity are 1,$\omega ,{\omega }^2$, then the roots of the equation $(x-1{)}^3+8=0$, are

• A. $-1,1-2\omega ,1-2{\omega }^2$
• B. $-1,1+2\omega ,1+2{\omega }^2$
• C. $-1,-1+2\omega ,-1-2{\omega }^2$
• D. $-1,1,-\omega$

Answer 1

Answer: (A)

$-1,1-2\omega ,1-2{\omega }^2$

${(x-1)}^3+8=0$
${(x-1)}^3=-8$
$\therefore (x-1)=-2.{\left(1\right)}^{1/3}$
$\therefore x-1=-2;x-1=-2w;x-1=-2w^2$
$\Rightarrow x=-1,x=1-2\omega ,x=1-2{\omega }^2$