Match List I with List II: \begin{array}{|l|l|} \hline \text{List I} & \text{List II} \\ \hline a. \text{The value of } \frac{1}{9}(1-w)(1-w^2)(1-w^4)(1-w^8) & I. 0 \\ b. \text{w} \left(1+w-w^2 \right)^7 & II. 1 \\ c. \text{The least positive integer n such that } \left(1+w^2\right)^n = \left(1+w^4\right)^n & III. -128 \\ d. \left(1+w+w^2\right) & IV. 3 \\ \hline \end{array} Choose the correct answer from the options given below:
Step-by-step Solution:
We will solve each part step by step. Part (a): Evaluating \( \frac{1}{9} (1 - w)(1 - w^2)(1 - w^4)(1 - w^8) \) We use the identity: \[ (1 - w)(1 - w^2)(1 - w^4)(1 - w^8) = 9 \] Thus, \[ \frac{1}{9} \times 9 = 1 \] So, \( A = II \). \[ \boxed{A - II} \] Part (b): Evaluating \( w(1 + w - w^2)^7 \) Given that: \[ 1 + w - w^2 = -2 \] \[ (1 + w - w^2)^7 = (-2)^7 = -128 \] Multiplying by \( w \), we get: \[ w(-128) = -128 \] Thus, \( B = III \). \[ \boxed{B - III} \] Part (c): Finding \( n \) in \( (1 + w^2)^n = (1 + w^4)^n \) Dividing both sides: \[ \left( \frac{1 + w^2}{1 + w^4} \right)^n = 1 \] Since \( \frac{1 + w^2}{1 + w^4} = w \), the smallest positive integer \( n \) that satisfies \( w^n = 1 \) is: \[ n = 3 \] Thus, \( C = IV \). \[ \boxed{C - IV} \] Part (d): Evaluating \( (1 + w + w^2) \) Using the known identity: \[ 1 + w + w^2 = 0 \] Thus, \( D = I \). \[ \boxed{D - I} \] Final Matching: \[ A - II, \quad B - III, \quad C - IV, \quad D - I \] This corresponds to option B. \[ \boxed{\text{Option B}} \]