Given the equation x6=64 Because equation degree is equal to = 6 - contains the even number 6 in the numerator, then the equation has two real roots. Get the root 6-th degree of the equation sides: We get: 6(1x+0)6=2 6(1x+0)6=−2 or x=2 x=−2 We get the answer: x = 2 We get the answer: x = -2 or x1=−2 x2=2
All other 4 root(s) is the complex numbers. do replacement: z=x then the equation will be the: z6=64 Any complex number can presented so: z=reip substitute to the equation r6e6ip=64 where r=2 - the magnitude of the complex number Substitute r: e6ip=1 Using Euler’s formula, we find roots for p isin(6p)+cos(6p)=1 so cos(6p)=1 and sin(6p)=0 then p=3πN 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=2 z3=−1−3i z4=−1+3i z5=1−3i z6=1+3i do backward replacement z=x x=z
The final answer: x1=−2 x2=2 x3=−1−3i x4=−1+3i x5=1−3i x6=1+3i