Question

If $1,\omega ,{\omega }^2$ are the cube roots of unity, then the value of
$(1+\omega )(1+{\omega }^2)(1+{\omega }^4)(1+{\omega }^8)$ is.

• A. $-1$
• B. $0$
• C. $1$
• D. $2$

Answer 1

The correct option is C $1$

Cube roots of unity satisy

$w^3=1,1+w+w^2=0$

$(1+w)(1+w^2)(1+w^4)(1+w^8)$

$=(1+w^2+w+w^3)(1+w^8+w^4+w^{12})$

$=(0+1)(1+w^2+w+(w^4{)}^3)$

$=\left(1\right)(0+1)$

$=1\left(1\right)=1$