Question

$1,\omega ,{\omega }^2$ are the cube roots of unity then
$(2-{\omega }^{})(2-{\omega }^2)(2-{\omega }^{13})(2-{\omega }^{17})$

• A. $36$
• B. $49$
• C. $40$
• D. $7^3$

Answer 1

Answer: (B)

$49$

$(2-\omega )(2-{\omega }^2)(2-{\omega }^{13})(2-{\omega }^{17})$

$=(2-\omega )(2-{\omega }^2)(2-{\omega }^{12}.\omega )(2-{\omega }^{15}.{\omega }^2)$

Since $\omega$ is a cube root of unity,

${\omega }^{12}=({\omega }^3{)}^4=1$

and

${\omega }^{15}=({\omega }^3{)}^5=1$

Hence,

$(2-\omega )(2-{\omega }^2)(2-{\omega }^{12}.\omega )(2-{\omega }^{15}.{\omega }^2)$

$=\left[\right(2-\omega \left)\right(2-{\omega }^2){]}^2$

$=[4-2(\omega +{\omega }^2)+\omega .{\omega }^2]$

We know that

$1+\omega +{\omega }^2=0$

$\therefore [4-2(\omega +{\omega }^2)+\omega .{\omega }^2]$

= $[4+2+1{]}^2=49$