Question

If \(1,\omega,\omega^2\) are the cube roots of unity, then \[ (2-\omega)^2(2-\omega^2)^2(2-\omega^{10})^2(2-\omega^{11})^2= \]

• A. \(-7^4\)
• B. \(7^4\)
• C. \(7^8\)
• D. \(-7^8\)

Hint:

For cube roots of unity, always use \(\omega^3=1\) and \(1+\omega+\omega^2=0\). These identities simplify higher powers and products quickly.

Answer 1

The Correct Option is B

Step 1: Use properties of cube roots of unity.
Since \(1,\omega,\omega^2\) are cube roots of unity, \[ \omega^3=1 \] Also, \[ 1+\omega+\omega^2=0 \] So, \[ \omega+\omega^2=-1 \]

Step 2: Simplify higher powers of \(\omega\).
\[ \omega^{10}=\omega^{9}\cdot \omega=(\omega^3)^3\omega=\omega \] and \[ \omega^{11}=\omega^{9}\cdot \omega^2=(\omega^3)^3\omega^2=\omega^2 \]

Step 3: Substitute these values.
\[ (2-\omega)^2(2-\omega^2)^2(2-\omega^{10})^2(2-\omega^{11})^2 \] \[ =(2-\omega)^2(2-\omega^2)^2(2-\omega)^2(2-\omega^2)^2 \] \[ =\left[(2-\omega)(2-\omega^2)\right]^4 \]

Step 4: Simplify the product.
\[ (2-\omega)(2-\omega^2) = 4-2\omega-2\omega^2+\omega\omega^2 \] Since \[ \omega+\omega^2=-1 \] and \[ \omega\omega^2=\omega^3=1 \] we get \[ (2-\omega)(2-\omega^2)=4-2(\omega+\omega^2)+1 \] \[ =4-2(-1)+1 \] \[ =7 \]

Step 5: Final conclusion.
Therefore, \[ \left[(2-\omega)(2-\omega^2)\right]^4=7^4 \] Hence, \[ \boxed{7^4} \]