Given the inequality:
$$x^{6} < 64$$
To solve this inequality, we must first solve the corresponding equation:
$$x^{6} = 64$$
Solve:
Given the equation
$$x^{6} = 64$$
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:
$$\sqrt[6]{x^{6}} = \sqrt[6]{64}$$
$$\sqrt[6]{x^{6}} = \left(-1\right) \sqrt[6]{64}$$
or
$$x = 2$$
$$x = -2$$
We get the answer: x = 2
We get the answer: x = -2
or
$$x_{1} = -2$$
$$x_{2} = 2$$
All other 4 root(s) is the complex numbers.
do replacement:
$$z = x$$
then the equation will be the:
$$z^{6} = 64$$
Any complex number can presented so:
$$z = r e^{i p}$$
substitute to the equation
$$r^{6} e^{6 i p} = 64$$
where
$$r = 2$$
- the magnitude of the complex number
Substitute r:
$$e^{6 i p} = 1$$
Using Euler’s formula, we find roots for p
$$i \sin{\left(6 p \right)} + \cos{\left(6 p \right)} = 1$$
so
$$\cos{\left(6 p \right)} = 1$$
and
$$\sin{\left(6 p \right)} = 0$$
then
$$p = \frac{\pi N}{3}$$
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:
$$z_{1} = -2$$
$$z_{2} = 2$$
$$z_{3} = -1 - \sqrt{3} i$$
$$z_{4} = -1 + \sqrt{3} i$$
$$z_{5} = 1 - \sqrt{3} i$$
$$z_{6} = 1 + \sqrt{3} i$$
do backward replacement
$$z = x$$
$$x = z$$
$$x_{1} = 2$$
$$x_{2} = -2$$
$$x_{1} = 2$$
$$x_{2} = -2$$
This roots
$$x_{2} = -2$$
$$x_{1} = 2$$
is the points with change the sign of the inequality expression.
First define with the sign to the leftmost point:
$$x_{0} < x_{2}$$
For example, let's take the point
$$x_{0} = x_{2} - \frac{1}{10}$$
=
$$-2 + - \frac{1}{10}$$
=
$$- \frac{21}{10}$$
substitute to the expression
$$x^{6} < 64$$
$$\left(- \frac{21}{10}\right)^{6} < 64$$
85766121
-------- < 64
1000000
but
85766121
-------- > 64
1000000
Then
$$x < -2$$
no execute
one of the solutions of our inequality is:
$$x > -2 \wedge x < 2$$
_____
/ \
-------ο-------ο-------
x2 x1