If is an imaginary cube root of unity, then the value of (1- + ² ) (1- ²+ ⁴ ) (1- ⁴+ ⁸ ) (2n factors) is

Step-by-step solution

Given, ³=1,1+ + ²=0 Now, (1- + ² ) (1- ²+ ⁴ ) (1- ⁴+ ⁸ ) . (1- ⁸+ ¹⁶ ) 2 n factors = (1- + ² ) (1- ²+ ) (1- + ² ) (1- ²+ ) .2 n factors [from Eq. (i)] =(- - ) (- ²- ² ) (- - ) (- ²- ² ) ... 2 n factors [from Eq. (i)] =(-2 ) (-2 ² ) (-2 ) (-2 ² ) 2 n factors =(-2)^ 2 n ( )^ n ( ² )^ n =(-1)^ 2 n (2)^ 2 n ^ 3 n =2^ 2 n ( ³ )^ n =2^ 2 n (1)^ n =2^ 2 n [from Eq. (i)]