Mister Exam
Other calculators
-
How to use it?
- Equation:
-
Equation 14/(x-3)=-7/9
-
Equation x^4+2*x^2-8=0
-
Equation 9x−9(x+1)=−2x
-
Equation x^4+4=0
- Express {x} in terms of y where:
- -17*x+1*y=-11
- 1*x+4*y=20
- -18*x-2*y=9
- -20*x-14*y=5
- Integral of d{x}:
-
x^3
- Derivative of:
-
x^3
- Limit of the function:
-
x^3
- Sum of series:
-
64
-
Identical expressions
- x^ three = sixty-four
- x cubed equally 64
- x to the power of three equally sixty minus four
- x3=64
- x³=64
- x to the power of 3=64
-
Similar expressions
- atan(x)-(1/x^3)=0
- 2x^3+4x^2-3x+6/4=0
- x^3+5*x^2-16x-80=0
- x^3+5x^2+14x=0
- (x-6)*(4x-6)=0
- 8sqrt(x^4-10*x^3+13)-8sqrt((x^2+3)^2-12)=0
x^3=64 equation
The teacher will be very surprised to see your correct solution 😉
The solution
Detail solution
Given the equation
$$x^{3} = 64$$
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:
$$\sqrt[3]{x^{3}} = \sqrt[3]{64}$$
or
$$x = 4$$
We get the answer: x = 4
All other 2 root(s) is the complex numbers.
do replacement:
$$z = x$$
then the equation will be the:
$$z^{3} = 64$$
Any complex number can presented so:
$$z = r e^{i p}$$
substitute to the equation
$$r^{3} e^{3 i p} = 64$$
where
$$r = 4$$
- 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:
$$z_{1} = 4$$
$$z_{2} = -2 - 2 \sqrt{3} i$$
$$z_{3} = -2 + 2 \sqrt{3} i$$
do backward replacement
$$z = x$$
$$x = z$$
The final answer:
$$x_{1} = 4$$
$$x_{2} = -2 - 2 \sqrt{3} i$$
$$x_{3} = -2 + 2 \sqrt{3} i$$
$$x^{3} = 64$$
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:
$$\sqrt[3]{x^{3}} = \sqrt[3]{64}$$
or
$$x = 4$$
We get the answer: x = 4
All other 2 root(s) is the complex numbers.
do replacement:
$$z = x$$
then the equation will be the:
$$z^{3} = 64$$
Any complex number can presented so:
$$z = r e^{i p}$$
substitute to the equation
$$r^{3} e^{3 i p} = 64$$
where
$$r = 4$$
- 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:
$$z_{1} = 4$$
$$z_{2} = -2 - 2 \sqrt{3} i$$
$$z_{3} = -2 + 2 \sqrt{3} i$$
do backward replacement
$$z = x$$
$$x = z$$
The final answer:
$$x_{1} = 4$$
$$x_{2} = -2 - 2 \sqrt{3} i$$
$$x_{3} = -2 + 2 \sqrt{3} i$$
Vieta's Theorem
it is reduced cubic equation
$$p x^{2} + q x + v + x^{3} = 0$$
where
$$p = \frac{b}{a}$$
$$p = 0$$
$$q = \frac{c}{a}$$
$$q = 0$$
$$v = \frac{d}{a}$$
$$v = -64$$
Vieta Formulas
$$x_{1} + x_{2} + x_{3} = - p$$
$$x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = q$$
$$x_{1} x_{2} x_{3} = v$$
$$x_{1} + x_{2} + x_{3} = 0$$
$$x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = 0$$
$$x_{1} x_{2} x_{3} = -64$$
$$p x^{2} + q x + v + x^{3} = 0$$
where
$$p = \frac{b}{a}$$
$$p = 0$$
$$q = \frac{c}{a}$$
$$q = 0$$
$$v = \frac{d}{a}$$
$$v = -64$$
Vieta Formulas
$$x_{1} + x_{2} + x_{3} = - p$$
$$x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = q$$
$$x_{1} x_{2} x_{3} = v$$
$$x_{1} + x_{2} + x_{3} = 0$$
$$x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = 0$$
$$x_{1} x_{2} x_{3} = -64$$
Rapid solution
[src]
x1 = 4
$$x_{1} = 4$$
___ x2 = -2 - 2*I*\/ 3
$$x_{2} = -2 - 2 \sqrt{3} i$$
___ x3 = -2 + 2*I*\/ 3
$$x_{3} = -2 + 2 \sqrt{3} i$$
x3 = -2 + 2*sqrt(3)*i
Sum and product of roots
[src]
sum
___ ___ 4 + -2 - 2*I*\/ 3 + -2 + 2*I*\/ 3
$$\left(4 + \left(-2 - 2 \sqrt{3} i\right)\right) + \left(-2 + 2 \sqrt{3} i\right)$$
=
0
$$0$$
product
/ ___\ / ___\ 4*\-2 - 2*I*\/ 3 /*\-2 + 2*I*\/ 3 /
$$4 \left(-2 - 2 \sqrt{3} i\right) \left(-2 + 2 \sqrt{3} i\right)$$
=
64
$$64$$
64
Numerical answer
[src]
x1 = 4.0
x2 = -2.0 - 3.46410161513775*i
x3 = -2.0 + 3.46410161513775*i
x3 = -2.0 + 3.46410161513775*i
The graph