Question 164272
i+1 =sqrt(2) [cos(pi/4) + isin(pi/4) ]
Now
(i+1)^6= (sqrt(2) )^6 [cos(pi/4) +isin(pi/4)]^6
By De Moivre's
(i+1)^6= 8 {cos (8pi/4) +isin(8pi/4) )
       = 8*1
       =8.