Mister Exam
Other calculators
-
How to use it?
- Equation:
-
Equation 12/x=3
-
Equation x^3=3
-
Equation 7x-4=x-16
-
Equation sin^24𝑥−sin4𝑥−2=0
- Express {x} in terms of y where:
- -6*x-11*y=-20
- -15*x+7*y=1
- -1*x-20*y=1
- -1*x-15*y=13
- Derivative of:
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x^3
- Integral of d{x}:
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x^3
- x^3
-
Identical expressions
- x^ three = three
- x cubed equally 3
- x to the power of three equally three
- x3=3
- x³=3
- x to the power of 3=3
-
Similar expressions
- x^3=0.34
x^3=3 equation
The teacher will be very surprised to see your correct solution 😉
The solution
Detail solution
Given the equation
$$x^{3} = 3$$
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:
$$\sqrt[3]{x^{3}} = \sqrt[3]{3}$$
or
$$x = \sqrt[3]{3}$$
Expand brackets in the right part
We get the answer: x = 3^(1/3)
All other 2 root(s) is the complex numbers.
do replacement:
$$z = x$$
then the equation will be the:
$$z^{3} = 3$$
Any complex number can presented so:
$$z = r e^{i p}$$
substitute to the equation
$$r^{3} e^{3 i p} = 3$$
where
$$r = \sqrt[3]{3}$$
- 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:
$$z_{1} = \sqrt[3]{3}$$
$$z_{2} = - \frac{\sqrt[3]{3}}{2} - \frac{3^{\frac{5}{6}} i}{2}$$
$$z_{3} = - \frac{\sqrt[3]{3}}{2} + \frac{3^{\frac{5}{6}} i}{2}$$
do backward replacement
$$z = x$$
$$x = z$$
The final answer:
$$x_{1} = \sqrt[3]{3}$$
$$x_{2} = - \frac{\sqrt[3]{3}}{2} - \frac{3^{\frac{5}{6}} i}{2}$$
$$x_{3} = - \frac{\sqrt[3]{3}}{2} + \frac{3^{\frac{5}{6}} i}{2}$$
$$x^{3} = 3$$
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:
$$\sqrt[3]{x^{3}} = \sqrt[3]{3}$$
or
$$x = \sqrt[3]{3}$$
Expand brackets in the right part
x = 3^1/3
We get the answer: x = 3^(1/3)
All other 2 root(s) is the complex numbers.
do replacement:
$$z = x$$
then the equation will be the:
$$z^{3} = 3$$
Any complex number can presented so:
$$z = r e^{i p}$$
substitute to the equation
$$r^{3} e^{3 i p} = 3$$
where
$$r = \sqrt[3]{3}$$
- 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:
$$z_{1} = \sqrt[3]{3}$$
$$z_{2} = - \frac{\sqrt[3]{3}}{2} - \frac{3^{\frac{5}{6}} i}{2}$$
$$z_{3} = - \frac{\sqrt[3]{3}}{2} + \frac{3^{\frac{5}{6}} i}{2}$$
do backward replacement
$$z = x$$
$$x = z$$
The final answer:
$$x_{1} = \sqrt[3]{3}$$
$$x_{2} = - \frac{\sqrt[3]{3}}{2} - \frac{3^{\frac{5}{6}} i}{2}$$
$$x_{3} = - \frac{\sqrt[3]{3}}{2} + \frac{3^{\frac{5}{6}} i}{2}$$
Vieta's Theorem
it is reduced cubic equation
$$p x^{2} + q x + v + x^{3} = 0$$
where
$$p = \frac{b}{a}$$
$$p = 0$$
$$q = \frac{c}{a}$$
$$q = 0$$
$$v = \frac{d}{a}$$
$$v = -3$$
Vieta Formulas
$$x_{1} + x_{2} + x_{3} = - p$$
$$x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = q$$
$$x_{1} x_{2} x_{3} = v$$
$$x_{1} + x_{2} + x_{3} = 0$$
$$x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = 0$$
$$x_{1} x_{2} x_{3} = -3$$
$$p x^{2} + q x + v + x^{3} = 0$$
where
$$p = \frac{b}{a}$$
$$p = 0$$
$$q = \frac{c}{a}$$
$$q = 0$$
$$v = \frac{d}{a}$$
$$v = -3$$
Vieta Formulas
$$x_{1} + x_{2} + x_{3} = - p$$
$$x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = q$$
$$x_{1} x_{2} x_{3} = v$$
$$x_{1} + x_{2} + x_{3} = 0$$
$$x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = 0$$
$$x_{1} x_{2} x_{3} = -3$$
Rapid solution
[src]
3 ___ x1 = \/ 3
$$x_{1} = \sqrt[3]{3}$$
3 ___ 5/6
\/ 3 I*3
x2 = - ----- - ------
2 2
$$x_{2} = - \frac{\sqrt[3]{3}}{2} - \frac{3^{\frac{5}{6}} i}{2}$$
3 ___ 5/6
\/ 3 I*3
x3 = - ----- + ------
2 2
$$x_{3} = - \frac{\sqrt[3]{3}}{2} + \frac{3^{\frac{5}{6}} i}{2}$$
x3 = -3^(1/3)/2 + 3^(5/6)*i/2
Sum and product of roots
[src]
sum
3 ___ 5/6 3 ___ 5/6
3 ___ \/ 3 I*3 \/ 3 I*3
\/ 3 + - ----- - ------ + - ----- + ------
2 2 2 2
$$\left(\sqrt[3]{3} + \left(- \frac{\sqrt[3]{3}}{2} - \frac{3^{\frac{5}{6}} i}{2}\right)\right) + \left(- \frac{\sqrt[3]{3}}{2} + \frac{3^{\frac{5}{6}} i}{2}\right)$$
=
0
$$0$$
product
/ 3 ___ 5/6\ / 3 ___ 5/6\
3 ___ | \/ 3 I*3 | | \/ 3 I*3 |
\/ 3 *|- ----- - ------|*|- ----- + ------|
\ 2 2 / \ 2 2 /
$$\sqrt[3]{3} \left(- \frac{\sqrt[3]{3}}{2} - \frac{3^{\frac{5}{6}} i}{2}\right) \left(- \frac{\sqrt[3]{3}}{2} + \frac{3^{\frac{5}{6}} i}{2}\right)$$
=
3
$$3$$
3
Numerical answer
[src]
x1 = -0.721124785153704 + 1.24902476648341*i
x2 = 1.44224957030741
x3 = -0.721124785153704 - 1.24902476648341*i
x3 = -0.721124785153704 - 1.24902476648341*i
The graph