If #w^3 = 1# where #w != 1# then what is the value of #(3+5w+3w^2)^2+(3+3w+5w^2)^2# ?

1 Answer
George C.
Sep 9, 2017

If #w != 1# satisfies #w^3=1# then:

#(3+5w+3w^2)^2+(3+3w+5w^2)^2 = -4#

Explanation:

Note that any cube root of unity is a root of:

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

So, assuming #w != 1#, then:

#w^2+w+1 = 0#

So:

#(3+5w+3w^2)^2+(3+3w+5w^2)^2#

#=(3(w^2+w+1)+2w)^2 + (3(w^2+w+1)+2w^2)^2#

#=(2w)^2+(2w^2)^2#

#=4(w^2+w^4)#

#=4(w^2+w)#

#=4(w^2+w+1-1)#

#=4(-1)#

#=-4#

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