Question

using de moivre's theoram show that x^7+x^4+x^3+1=0​

Answer 1

Answer:

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Step-by-step explanation:

x8−1)=(x−1)(x7+x6+x5+x4+x3+x2+x+1

x8−1=⟹x8=1|1|=1arg of 1=0

8 distinct 8th roots of unity can be found by

wk=cos(2kπ8)+isin(2kπ8)where k=0,1,2,⋯,7

⟹wk=cos(kπ4)+isin(kπ4)

w0=1

w1=12–√+12–√i

w2=i

w3=−12–√+12–√i

w4=−1

w5=−12–√−12–√i

w6=−i

w7=12–√−12–√i

We can leave 1, the first root. Remaining 7 are the roots of the given polynomial.