## How to solve x3=−1? I got following: x3=−1 x=(−1)13 x=(−1)12(−1)16=i(−1)16... algebra-precalculus

x3+1=0(x+1)(x2x+1)=0
If x+1=0,x=1
else x2x+1=0 then x=1±1421=1±3i2 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,...n1. 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+n1, where r is any integer), the reason is explained below.
(1)Using Repeated Root Theorem, let f(x)=xn+R where n>1 and R0x0
So, dfdx=nxn1,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(ab)πn=1=e2kπi where k is any integer.
ab=knab(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=0cosπ3+isinπ3=1+i32
m=1cosπ+isinπ=1
m=2cos5π3+isin5π3=1i32