Question

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

• A. $2,2\omega ,2{\omega }^2$
• B. $0,2(1-\omega ),2(1-{\omega }^2)$
• C. $2,0,\omega$
• D. $-2,-2\omega ,-2{\omega }^2$

Answer 1

The correct option is D $-2,-2\omega ,-2{\omega }^2$

$(x-2{)}^3+8=0$ $(x-2{)}^3=-8$

$\therefore (x-2)=(-2)(1{)}^{1/3}$

$1,\omega ,{\omega }^2$ are cube roots of unity

$\therefore x-2=-2$ $\therefore x=0$

OR x-2=$-2\omega$ $x=2-2\omega$

OR x-2 = $-2{\omega }^2$ $x=2-2{\omega }^2$