Question

If w is a complex cube root of 1 then show that :
(3+3w+5w^2)^6 = 64​

Answer 1

Answer:

1+w+w^2=0

so that 3+3w+5w^2=3(1+w+w^2)+2w^2=3×0+2w^2=2w^2

(3+3w+5w^2)^6=(2w^2)^6=64w^12

we know that w^3=1 ,so that w^12=1;

therefore 64w^12=64

Answer 2

Step-by-step explanation:

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