Question

Arrange the following values in ascending order:
I) $(1+\omega )(1+{\omega }^2)(1+{\omega }^4)(1+{\omega }^8)$
II) $(1-\omega +{\omega }^2{)}^7+(1+\omega -{\omega }^2{)}^7$
III) $(1-\omega )(1-{\omega }^2)(1-{\omega }^4)(1-{\omega }^8)$
IV) $(1-\omega +{\omega }^2)(1-{\omega }^2+{\omega }^4)$

• A. I,IV,III,II
• B. I,II,III,IV
• C. I,III,IV,II
• D. IV,III,II,I

Answer 1

Answer: (A)

I,IV,III,II

Using two formulae we can solve the problem , i.e.

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

2) ${\omega }^3=1$

I) $(1+\omega )(1+{\omega }^2)(1+{\omega }^4)(1+{\omega }^8)$

$=(-{\omega }^2)(-\omega )(1+\omega )(1+{\omega }^2)$

$=(-{\omega }^2)(-\omega )(-{\omega }^2)(-\omega )$

$={\omega }^6=1$

II) ${(1-\omega +{\omega }^2)}^7+{(1+\omega -{\omega }^2)}^7=128$

III) $(1-\omega )(1-{\omega }^2)(1-{\omega }^4)(1-{\omega }^8)$

$=(1-\omega )(1-{\omega }^2)(1-\omega )(1-{\omega }^2)$

$=9$

IV) $(1-\omega +{\omega }^2)(1-{\omega }^2+{\omega }^4)$

$=(-\omega -\omega )(-{\omega }^2-{\omega }^2)$

$=4{\omega }^3=4$

Hence, the correct sequence is I, IV, III, II.