Question

# If $\omega$ is an imaginary cube root of unity, then ${{\left( 1+\omega -{{\omega }^{2}} \right)}^{7}}$ equals(a) $128\omega$ (b) $-128\omega$ (c) $128{{\omega }^{2}}$ (d) $-128{{\omega }^{2}}$

Hint: The sum of $1,\omega$ and ${{\omega }^{2}}$ is equal to 0 where $1,\omega$ and ${{\omega }^{2}}$ are the cube roots of unity . Also, ${{\omega }^{3}}=1$.

Before proceeding with the question, we must know some properties which are related to the cube roots of unity i.e. $1,\omega ,{{\omega }^{2}}$ which will be used to solve this question. These properties are,
$1+\omega +{{\omega }^{2}}=0..........\left( 1 \right)$
${{\omega }^{3n}}=1..........\left( 2 \right)$, where $n$ is an integer.
In this question, we have to find the value of ${{\left( 1+\omega -{{\omega }^{2}} \right)}^{7}}$. From equation $\left( 1 \right)$, we have,
$1+\omega +{{\omega }^{2}}=0$
Hence, we can also write $1+\omega =-{{\omega }^{2}}.............\left( 4 \right)$
Substituting $1+\omega =-{{\omega }^{2}}$ from equation $\left( 4 \right)$ in equation $\left( 2 \right)$, we get,
\begin{align} & {{\left( 1+\omega -{{\omega }^{2}} \right)}^{7}}={{\left( -{{\omega }^{2}}-{{\omega }^{2}} \right)}^{7}} \\ & \Rightarrow {{\left( 1+\omega -{{\omega }^{2}} \right)}^{7}}={{\left( -2{{\omega }^{2}} \right)}^{7}} \\ & \Rightarrow {{\left( 1+\omega -{{\omega }^{2}} \right)}^{7}}={{\left( -2 \right)}^{7}}{{\left( {{\omega }^{2}} \right)}^{7}} \\ & \Rightarrow {{\left( 1+\omega -{{\omega }^{2}} \right)}^{7}}=-128{{\omega }^{14}} \\ & \Rightarrow {{\left( 1+\omega -{{\omega }^{2}} \right)}^{7}}=-128{{\omega }^{12+2}} \\ & \Rightarrow {{\left( 1+\omega -{{\omega }^{2}} \right)}^{7}}=-128{{\omega }^{12}}{{\omega }^{2}} \\ & \Rightarrow {{\left( 1+\omega -{{\omega }^{2}} \right)}^{7}}=-128{{\omega }^{3\left( 4 \right)}}{{\omega }^{2}}...............\left( 5 \right) \\ \end{align}
From equation $\left( 2 \right)$, we have ${{\omega }^{3n}}=1$ where $n$ is an integer. Since $4$ is an integer, we can substitute $n=4$ in equation $\left( 2 \right)$. So, substituting $n=4$ in equation $\left( 2 \right)$, we get,
${{\omega }^{3(4)}}=1.......\left( 6 \right)$
From $\left( 6 \right)$, we have ${{\omega }^{3(4)}}=1$. Substituting ${{\omega }^{3(4)}}=1$ from equation $\left( 6 \right)$ in equation $\left( 5 \right)$, we get,
\begin{align} & {{\left( 1+\omega -{{\omega }^{2}} \right)}^{7}}=-128\left( 1 \right){{\omega }^{2}} \\ & \Rightarrow {{\left( 1+\omega -{{\omega }^{2}} \right)}^{7}}=-128{{\omega }^{2}} \\ \end{align}

Note: In this question, it was easier to think of writing ${{\omega }^{14}}$ as ${{\omega }^{12+2}}$ where $12$ is a multiple of $3$ because $14$ is a comparatively smaller number and it is easier to express $14$ in the form of a multiple of $3$. But if we get a larger number, it is difficult to convert that number directly in the form of the multiple of $3$. So, in that case, we will divide that number by long division method to find the divisor, quotient and remainder. Then we express that number in the form of the multiple of $3$ using the formula, $number=\left( divisor \right)\times \left( quotient \right)+remainder$, where the $divisor=3$. For example if we have a number $149$ in the power of $\omega$, dividing $149$ by $3$ using a long division method, we will get $divisor=3,quotient=49,remainder=2$. Hence, we can express $149$ as $149=\left( 3 \right)\left( 49 \right)+2$. Therefore, we expressed $149$ in the form of multiple of $3$.