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x^6+64=0 equation

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Numerical solution:

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The solution

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 6         
x  + 64 = 0
x6+64=0x^{6} + 64 = 0
Simplify
Detail solution
Given the equation
x6+64=0x^{6} + 64 = 0
Because equation degree is equal to = 6 and the free term = -64 < 0,
so the real solutions of the equation d'not exist

All other 6 root(s) is the complex numbers.
do replacement:
z=xz = x
then the equation will be the:
z6=−64z^{6} = -64
Any complex number can presented so:
z=reipz = r e^{i p}
substitute to the equation
r6e6ip=−64r^{6} e^{6 i p} = -64
where
r=2r = 2
- the magnitude of the complex number
Substitute r:
e6ip=−1e^{6 i p} = -1
Using Euler’s formula, we find roots for p
isin⁡(6p)+cos⁡(6p)=−1i \sin{\left(6 p \right)} + \cos{\left(6 p \right)} = -1
so
cos⁡(6p)=−1\cos{\left(6 p \right)} = -1
and
sin⁡(6p)=0\sin{\left(6 p \right)} = 0
then
p=πN3+π6p = \frac{\pi N}{3} + \frac{\pi}{6}
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=−2iz_{1} = - 2 i
z2=2iz_{2} = 2 i
z3=−3−iz_{3} = - \sqrt{3} - i
z4=−3+iz_{4} = - \sqrt{3} + i
z5=3−iz_{5} = \sqrt{3} - i
z6=3+iz_{6} = \sqrt{3} + i
do backward replacement
z=xz = x
x=zx = z

The final answer:
x1=−2ix_{1} = - 2 i
x2=2ix_{2} = 2 i
x3=−3−ix_{3} = - \sqrt{3} - i
x4=−3+ix_{4} = - \sqrt{3} + i
x5=3−ix_{5} = \sqrt{3} - i
x6=3+ix_{6} = \sqrt{3} + i
Sum and product of roots [src] 
sum
                    ___         ___     ___             ___
-2*I + 2*I + -I - \/ 3  + I - \/ 3  + \/ 3  - I + I + \/ 3
((3−i)+(((−3−i)+(−2i+2i))+(−3+i)))+(3+i)\left(\left(\sqrt{3} - i\right) + \left(\left(\left(- \sqrt{3} - i\right) + \left(- 2 i + 2 i\right)\right) + \left(- \sqrt{3} + i\right)\right)\right) + \left(\sqrt{3} + i\right) 
=
0
00 
product
         /       ___\ /      ___\ /  ___    \ /      ___\
-2*I*2*I*\-I - \/ 3 /*\I - \/ 3 /*\\/ 3  - I/*\I + \/ 3 /
−2i2i(−3−i)(−3+i)(3−i)(3+i)- 2 i 2 i \left(- \sqrt{3} - i\right) \left(- \sqrt{3} + i\right) \left(\sqrt{3} - i\right) \left(\sqrt{3} + i\right) 
=
64
6464 
64
Rapid solution [src] 
x1 = -2*I
x1=−2ix_{1} = - 2 i 
x2 = 2*I
x2=2ix_{2} = 2 i 
            ___
x3 = -I - \/ 3
x3=−3−ix_{3} = - \sqrt{3} - i 
           ___
x4 = I - \/ 3
x4=−3+ix_{4} = - \sqrt{3} + i 
       ___    
x5 = \/ 3  - I
x5=3−ix_{5} = \sqrt{3} - i 
           ___
x6 = I + \/ 3
x6=3+ix_{6} = \sqrt{3} + i 
x6 = sqrt(3) + i
Numerical answer [src] 
x1 = -1.73205080756888 + 1.0*i
x2 = -1.73205080756888 - 1.0*i
x3 = 2.0*i
x4 = -2.0*i
x5 = 1.73205080756888 + 1.0*i
x6 = 1.73205080756888 - 1.0*i
x6 = 1.73205080756888 - 1.0*i
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