Question

If \[\omega\] is an imaginary cube root of unity, then \[{{\text{(1+}\omega -{{\omega }^{2}}\text{)}}^{7}}\] equals-

Answer 1

Hint: We have the expression, \[{{\text{(1+}\omega -{{\omega }^{2}}\text{)}}^{7}}\] . We need to find the value of this expression. First make the expression in terms of \[{{\omega }^{2}}\] . We know the property, \[\text{1+}\omega +{{\omega }^{2}}=0\]. Using this property, put the value of \[\text{1+}\omega =-{{\omega }^{2}}\] and solve it further.

Complete step by step answer:
We know that,
\[\text{1+}\omega +{{\omega }^{2}}=0\]
\[\Rightarrow \text{1+}\omega =-{{\omega }^{2}}\] ……..eq(i)
Putting eq(i) in \[{{\text{(1+}\omega -{{\omega }^{2}}\text{)}}^{7}}\]
 we get,
\[{{\text{(1+}\omega -{{\omega }^{2}}\text{)}}^{7}}\text{=(}-{{\omega }^{2}}-{{\omega }^{2}}{{\text{)}}^{7}}\]
\[={{(-2{{\omega }^{2}})}^{7}}=-128{{\omega }^{2.7}}=-128{{\omega }^{14}}\] ………eq(ii)
We know that, \[{{\omega }^{3n+2}}={{\omega }^{2}}\]
and \[{{\omega }^{3n}}=1\]
Thus we get,
\[{{\omega }^{14}}={{\omega }^{3.4}}\times {{\omega }^{2}}=1\times {{\omega }^{2}}={{\omega }^{2}}\] …….eq(iii)
Putting eq.(iii) in eq.(ii),we get
\[\text{-128}\times {{\omega }^{14}}=-128\times {{\omega }^{2}}=-128{{\omega }^{2}}\]

Hence, \[{{\text{(1+}\omega -{{\omega }^{2}}\text{)}}^{7}}=-128{{\omega }^{2}}\] .

Note: In this type of question, one can think to expand the given expression directly after putting the values of \[\omega \] and \[{{\omega }^{2}}\] . And then it will become complex to solve further and a lot of time can get wasted. To overcome this situation, first, try to make the expression in a single cube root of unity using its property and then expand. After expansion, using the properties of cube roots we can easily solve this question and conclude the answer.