x3+1=0⟹(x+1)(x2−x+1)=0
If x+1=0,x=−1
else x2−x+1=0 then x=1±1−4√2⋅1=1±3√i2 using the well known Quadratic Formula.
Alternatively, using Euler's identity and Euler's formula like other two solutions,
for xn=−1=ei(2m+1)π as eiπ=−1 where m any any integer, n is a natural number .
We know n-degree equation has exactly n roots, so the roots of xn+1=0
are ei(2m+1)πn=cos(2m+1)πn+isin(2m+1)πn where m=0,1,2,...n−1. It's just customary, not mandatory that we have defined m to assume this range of values, in fact it can assume any n in-congruent values (for example, consecutive values like r,r+1,...,r+n−1, where r is any integer), the reason is explained below.
(1)Using Repeated Root Theorem, let f(x)=xn+R where n>1 and R≠0⟹x≠0
So, dfdx=nxn−1,dfdx=0 does not have any non-zero root.
Clearly, f(x)=xn+R can not have any repeated root unless R=0
(2)Let ei(2a+1)πn=ei(2b+1)πn
⟹e2i(a−b)πn=1=e2kπi where k is any integer.
⟹a−b=kn⟹a≡b(modn), so any n in-congruent values of m will give us essentially the same set of n distinct solutions.
Here n=3 so, let's take m=0,1,2.
m=0⟹cosπ3+isinπ3=1+i3√2
m=1⟹cosπ+isinπ=−1
m=2⟹cos5π3+isin5π3=1−i3√2
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