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# 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) $-128{{\omega }^{2}}$  Answer Verified
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}

So,the answer is option (d)

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 $3$.
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Cube Root of Unity  Cube Root  Cube Root of 729  Cube Root of 1728  Cube Root of 2197  Cube Root of 512  Cube Root of 343  Cube Root of 216  Cube Root of 64  Cube Root of 4  