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x^3=0.34

x^3=0.34 equation

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Numerical solution:

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The solution

You have entered [src] 
 3   17
x  = --
     50
x3=1750x^{3} = \frac{17}{50}
Simplify
Detail solution
Given the equation
x3=1750x^{3} = \frac{17}{50}
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:
(1x+0)33=17503\sqrt[3]{\left(1 x + 0\right)^{3}} = \sqrt[3]{\frac{17}{50}}
or
x=340310x = \frac{\sqrt[3]{340}}{10}
Expand brackets in the right part
x = 340^1/3/10

We get the answer: x = 340^(1/3)/10

All other 2 root(s) is the complex numbers.
do replacement:
z=xz = x
then the equation will be the:
z3=1750z^{3} = \frac{17}{50}
Any complex number can presented so:
z=reipz = r e^{i p}
substitute to the equation
r3e3ip=1750r^{3} e^{3 i p} = \frac{17}{50}
where
r=340310r = \frac{\sqrt[3]{340}}{10}
- the magnitude of the complex number
Substitute r:
e3ip=1e^{3 i p} = 1
Using Euler’s formula, we find roots for p
isin⁡(3p)+cos⁡(3p)=1i \sin{\left(3 p \right)} + \cos{\left(3 p \right)} = 1
so
cos⁡(3p)=1\cos{\left(3 p \right)} = 1
and
sin⁡(3p)=0\sin{\left(3 p \right)} = 0
then
p=2πN3p = \frac{2 \pi N}{3}
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=340310z_{1} = \frac{\sqrt[3]{340}}{10}
z2=−340320−3⋅3403i20z_{2} = - \frac{\sqrt[3]{340}}{20} - \frac{\sqrt{3} \cdot \sqrt[3]{340} i}{20}
z3=−340320+3⋅3403i20z_{3} = - \frac{\sqrt[3]{340}}{20} + \frac{\sqrt{3} \cdot \sqrt[3]{340} i}{20}
do backward replacement
z=xz = x
x=zx = z

The final answer:
x1=340310x_{1} = \frac{\sqrt[3]{340}}{10}
x2=−340320−3⋅3403i20x_{2} = - \frac{\sqrt[3]{340}}{20} - \frac{\sqrt{3} \cdot \sqrt[3]{340} i}{20}
x3=−340320+3⋅3403i20x_{3} = - \frac{\sqrt[3]{340}}{20} + \frac{\sqrt{3} \cdot \sqrt[3]{340} i}{20}
Vieta's Theorem
it is reduced cubic equation
px2+qx+v+x3=0p x^{2} + q x + v + x^{3} = 0
where
p=bap = \frac{b}{a}
p=0p = 0
q=caq = \frac{c}{a}
q=0q = 0
v=dav = \frac{d}{a}
v=−1750v = - \frac{17}{50}
Vieta Formulas
x1+x2+x3=−px_{1} + x_{2} + x_{3} = - p
x1x2+x1x3+x2x3=qx_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = q
x1x2x3=vx_{1} x_{2} x_{3} = v
x1+x2+x3=0x_{1} + x_{2} + x_{3} = 0
x1x2+x1x3+x2x3=0x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = 0
x1x2x3=−1750x_{1} x_{2} x_{3} = - \frac{17}{50}
The graph
-12.5-10.0-7.5-5.0-2.50.02.55.07.510.012.515.0-20002000
Plot the graph f1(x)
Plot the graph f2(x)
Sum and product of roots [src] 
sum
    3 _____     3 _____       ___ 3 _____     3 _____       ___ 3 _____
    \/ 340      \/ 340    I*\/ 3 *\/ 340      \/ 340    I*\/ 3 *\/ 340 
0 + ------- + - ------- - --------------- + - ------- + ---------------
       10          20            20              20            20
((0+340310)−(340320+3⋅3403i20))−(340320−3⋅3403i20)\left(\left(0 + \frac{\sqrt[3]{340}}{10}\right) - \left(\frac{\sqrt[3]{340}}{20} + \frac{\sqrt{3} \cdot \sqrt[3]{340} i}{20}\right)\right) - \left(\frac{\sqrt[3]{340}}{20} - \frac{\sqrt{3} \cdot \sqrt[3]{340} i}{20}\right) 
=
0
00 
product
  3 _____ /  3 _____       ___ 3 _____\ /  3 _____       ___ 3 _____\
  \/ 340  |  \/ 340    I*\/ 3 *\/ 340 | |  \/ 340    I*\/ 3 *\/ 340 |
1*-------*|- ------- - ---------------|*|- ------- + ---------------|
     10   \     20            20      / \     20            20      /
1⋅340310(−340320−3⋅3403i20)(−340320+3⋅3403i20)1 \cdot \frac{\sqrt[3]{340}}{10} \left(- \frac{\sqrt[3]{340}}{20} - \frac{\sqrt{3} \cdot \sqrt[3]{340} i}{20}\right) \left(- \frac{\sqrt[3]{340}}{20} + \frac{\sqrt{3} \cdot \sqrt[3]{340} i}{20}\right) 
=
17
--
50
1750\frac{17}{50} 
17/50
Rapid solution [src] 
     3 _____
     \/ 340 
x1 = -------
        10
x1=340310x_{1} = \frac{\sqrt[3]{340}}{10}
Simplify 
       3 _____       ___ 3 _____
       \/ 340    I*\/ 3 *\/ 340 
x2 = - ------- - ---------------
          20            20
x2=−340320−3⋅3403i20x_{2} = - \frac{\sqrt[3]{340}}{20} - \frac{\sqrt{3} \cdot \sqrt[3]{340} i}{20}
Simplify 
       3 _____       ___ 3 _____
       \/ 340    I*\/ 3 *\/ 340 
x3 = - ------- + ---------------
          20            20
x3=−340320+3⋅3403i20x_{3} = - \frac{\sqrt[3]{340}}{20} + \frac{\sqrt{3} \cdot \sqrt[3]{340} i}{20}
Simplify
Numerical answer [src] 
x1 = -0.348976602345444 - 0.60444520591507*i
x2 = -0.348976602345444 + 0.60444520591507*i
x3 = 0.697953204690889
x3 = 0.697953204690889
The graph
x^3=0.34 equation
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