If #omega#(not equal to #1#) is a cube root of unity,and #(1+omega)^7 = A+Bomega#. Then find #(A,B)# ?

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Answer 1 · 2018-03-27T07:39:41

#A=1# and #B=1#

Note that:

#0 = omega^3-1 = (omega-1)(omega^2+omega+1)#

So since #omega != 1#, it must satisfy:

#omega^2+omega+1 = 0#

So:

#(1 + omega)^6 = (-omega^2)^6 = omega^12 = (omega^3)^4 = 1#

So:

#(1+omega)^7 = (1+omega)^6(1+omega) = 1+omega#

So #A=1# and #B=1#