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x^4-81=0 equation

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Numerical solution:

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The solution

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 4         
x  - 81 = 0
x481=0x^{4} - 81 = 0
Detail solution
Given the equation
x481=0x^{4} - 81 = 0
Because equation degree is equal to = 4 - contains the even number 4 in the numerator, then
the equation has two real roots.
Get the root 4-th degree of the equation sides:
We get:
x44=814\sqrt[4]{x^{4}} = \sqrt[4]{81}
x44=(1)814\sqrt[4]{x^{4}} = \left(-1\right) \sqrt[4]{81}
or
x=3x = 3
x=3x = -3
We get the answer: x = 3
We get the answer: x = -3
or
x1=3x_{1} = -3
x2=3x_{2} = 3

All other 2 root(s) is the complex numbers.
do replacement:
z=xz = x
then the equation will be the:
z4=81z^{4} = 81
Any complex number can presented so:
z=reipz = r e^{i p}
substitute to the equation
r4e4ip=81r^{4} e^{4 i p} = 81
where
r=3r = 3
- the magnitude of the complex number
Substitute r:
e4ip=1e^{4 i p} = 1
Using Euler’s formula, we find roots for p
isin(4p)+cos(4p)=1i \sin{\left(4 p \right)} + \cos{\left(4 p \right)} = 1
so
cos(4p)=1\cos{\left(4 p \right)} = 1
and
sin(4p)=0\sin{\left(4 p \right)} = 0
then
p=πN2p = \frac{\pi N}{2}
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=3z_{1} = -3
z2=3z_{2} = 3
z3=3iz_{3} = - 3 i
z4=3iz_{4} = 3 i
do backward replacement
z=xz = x
x=zx = z

The final answer:
x1=3x_{1} = -3
x2=3x_{2} = 3
x3=3ix_{3} = - 3 i
x4=3ix_{4} = 3 i
Sum and product of roots [src]
sum
-3 + 3 - 3*I + 3*I
((3+3)3i)+3i\left(\left(-3 + 3\right) - 3 i\right) + 3 i
=
0
00
product
-3*3*-3*I*3*I
3i9(3i)3 i - 9 \left(- 3 i\right)
=
-81
81-81
-81
Rapid solution [src]
x1 = -3
x1=3x_{1} = -3
x2 = 3
x2=3x_{2} = 3
x3 = -3*I
x3=3ix_{3} = - 3 i
x4 = 3*I
x4=3ix_{4} = 3 i
x4 = 3*i
Numerical answer [src]
x1 = -3.0
x2 = 3.0*i
x3 = 3.0
x4 = -3.0*i
x4 = -3.0*i