x^3-8=0 equation
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The solution
Detail solution
Given the equation
$$x^{3} - 8 = 0$$
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:
$$\sqrt[3]{x^{3}} = \sqrt[3]{8}$$
or
$$x = 2$$
We get the answer: x = 2
All other 2 root(s) is the complex numbers.
do replacement:
$$z = x$$
then the equation will be the:
$$z^{3} = 8$$
Any complex number can presented so:
$$z = r e^{i p}$$
substitute to the equation
$$r^{3} e^{3 i p} = 8$$
where
$$r = 2$$
- the magnitude of the complex number
Substitute r:
$$e^{3 i p} = 1$$
Using Euler’s formula, we find roots for p
$$i \sin{\left(3 p \right)} + \cos{\left(3 p \right)} = 1$$
so
$$\cos{\left(3 p \right)} = 1$$
and
$$\sin{\left(3 p \right)} = 0$$
then
$$p = \frac{2 \pi N}{3}$$
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:
$$z_{1} = 2$$
$$z_{2} = -1 - \sqrt{3} i$$
$$z_{3} = -1 + \sqrt{3} i$$
do backward replacement
$$z = x$$
$$x = z$$
The final answer:
$$x_{1} = 2$$
$$x_{2} = -1 - \sqrt{3} i$$
$$x_{3} = -1 + \sqrt{3} i$$
Vieta's Theorem
it is reduced cubic equation
$$p x^{2} + q x + v + x^{3} = 0$$
where
$$p = \frac{b}{a}$$
$$p = 0$$
$$q = \frac{c}{a}$$
$$q = 0$$
$$v = \frac{d}{a}$$
$$v = -8$$
Vieta Formulas
$$x_{1} + x_{2} + x_{3} = - p$$
$$x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = q$$
$$x_{1} x_{2} x_{3} = v$$
$$x_{1} + x_{2} + x_{3} = 0$$
$$x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = 0$$
$$x_{1} x_{2} x_{3} = -8$$
Sum and product of roots
[src]
___ ___
2 + -1 - I*\/ 3 + -1 + I*\/ 3
$$\left(2 + \left(-1 - \sqrt{3} i\right)\right) + \left(-1 + \sqrt{3} i\right)$$
$$0$$
/ ___\ / ___\
2*\-1 - I*\/ 3 /*\-1 + I*\/ 3 /
$$2 \left(-1 - \sqrt{3} i\right) \left(-1 + \sqrt{3} i\right)$$
$$8$$
$$x_{1} = 2$$
$$x_{2} = -1 - \sqrt{3} i$$
$$x_{3} = -1 + \sqrt{3} i$$
x2 = -1.0 - 1.73205080756888*i
x3 = -1.0 + 1.73205080756888*i
x3 = -1.0 + 1.73205080756888*i