Given the equation (x−1)3=−8 Because equation degree is equal to = 3 - does not contain even numbers in the numerator, then the equation has single real root. Get the root 3-th degree of the equation sides: We get: 3(x−1)3=3−8 or x−1=23−1 Expand brackets in the right part
-1 + x = -2*1^1/3
Move free summands (without x) from left part to right part, we given: x=1+23−1 We get the answer: x = 1 + 2*(-1)^(1/3)
All other 2 root(s) is the complex numbers. do replacement: z=x−1 then the equation will be the: z3=−8 Any complex number can presented so: z=reip substitute to the equation r3e3ip=−8 where r=2 - the magnitude of the complex number Substitute r: e3ip=−1 Using Euler’s formula, we find roots for p isin(3p)+cos(3p)=−1 so cos(3p)=−1 and sin(3p)=0 then p=32πN+3π where N=0,1,2,3,... Looping through the values of N and substituting p into the formula for z Consequently, the solution will be for z: z1=−2 z2=1−3i z3=1+3i do backward replacement z=x−1 x=z+1