Mister Exam
Other calculators
-
How to use it?
- Equation:
-
Equation x^3=27
-
Equation x^4+2*x^2-8=0
-
Equation cos(x)=2/3
-
Equation (x-11)^4=(x+3)^4
- Express {x} in terms of y where:
- 6*x+17*y=3
- 1*x+4*y=20
- 20*x+7*y=-9
- -17*x+18*y=10
- Derivative of:
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x^3
- Integral of d{x}:
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x^3
- x^3
- Sum of series:
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27
-
Identical expressions
- x^ three = twenty-seven
- x cubed equally 27
- x to the power of three equally twenty minus seven
- x3=27
- x³=27
- x to the power of 3=27
-
Similar expressions
- 8sqrt(x^4-10*x^3+13)-8sqrt((x^2+3)^2-12)=0
- 10^5/(220^2*0,52)*(7,4*10^-3*0,52+(0,116*10^-3)*0,82)*2*35=x
- -68(x^3)sin(17x^4)=0
- x^3+y^3=x^2+42*x*y+y^2
- 6x^3-24x=0
x^3=27 equation
The teacher will be very surprised to see your correct solution 😉
The solution
Detail solution
Given the equation
$$x^{3} = 27$$
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:
$$\sqrt[3]{x^{3}} = \sqrt[3]{27}$$
or
$$x = 3$$
We get the answer: x = 3
All other 2 root(s) is the complex numbers.
do replacement:
$$z = x$$
then the equation will be the:
$$z^{3} = 27$$
Any complex number can presented so:
$$z = r e^{i p}$$
substitute to the equation
$$r^{3} e^{3 i p} = 27$$
where
$$r = 3$$
- 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:
$$z_{1} = 3$$
$$z_{2} = - \frac{3}{2} - \frac{3 \sqrt{3} i}{2}$$
$$z_{3} = - \frac{3}{2} + \frac{3 \sqrt{3} i}{2}$$
do backward replacement
$$z = x$$
$$x = z$$
The final answer:
$$x_{1} = 3$$
$$x_{2} = - \frac{3}{2} - \frac{3 \sqrt{3} i}{2}$$
$$x_{3} = - \frac{3}{2} + \frac{3 \sqrt{3} i}{2}$$
$$x^{3} = 27$$
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:
$$\sqrt[3]{x^{3}} = \sqrt[3]{27}$$
or
$$x = 3$$
We get the answer: x = 3
All other 2 root(s) is the complex numbers.
do replacement:
$$z = x$$
then the equation will be the:
$$z^{3} = 27$$
Any complex number can presented so:
$$z = r e^{i p}$$
substitute to the equation
$$r^{3} e^{3 i p} = 27$$
where
$$r = 3$$
- 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:
$$z_{1} = 3$$
$$z_{2} = - \frac{3}{2} - \frac{3 \sqrt{3} i}{2}$$
$$z_{3} = - \frac{3}{2} + \frac{3 \sqrt{3} i}{2}$$
do backward replacement
$$z = x$$
$$x = z$$
The final answer:
$$x_{1} = 3$$
$$x_{2} = - \frac{3}{2} - \frac{3 \sqrt{3} i}{2}$$
$$x_{3} = - \frac{3}{2} + \frac{3 \sqrt{3} i}{2}$$
Vieta's Theorem
it is reduced cubic equation
$$p x^{2} + q x + v + x^{3} = 0$$
where
$$p = \frac{b}{a}$$
$$p = 0$$
$$q = \frac{c}{a}$$
$$q = 0$$
$$v = \frac{d}{a}$$
$$v = -27$$
Vieta Formulas
$$x_{1} + x_{2} + x_{3} = - p$$
$$x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = q$$
$$x_{1} x_{2} x_{3} = v$$
$$x_{1} + x_{2} + x_{3} = 0$$
$$x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = 0$$
$$x_{1} x_{2} x_{3} = -27$$
$$p x^{2} + q x + v + x^{3} = 0$$
where
$$p = \frac{b}{a}$$
$$p = 0$$
$$q = \frac{c}{a}$$
$$q = 0$$
$$v = \frac{d}{a}$$
$$v = -27$$
Vieta Formulas
$$x_{1} + x_{2} + x_{3} = - p$$
$$x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = q$$
$$x_{1} x_{2} x_{3} = v$$
$$x_{1} + x_{2} + x_{3} = 0$$
$$x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = 0$$
$$x_{1} x_{2} x_{3} = -27$$
Rapid solution
[src]
x1 = 3
$$x_{1} = 3$$
___
3 3*I*\/ 3
x2 = - - - ---------
2 2
$$x_{2} = - \frac{3}{2} - \frac{3 \sqrt{3} i}{2}$$
___
3 3*I*\/ 3
x3 = - - + ---------
2 2
$$x_{3} = - \frac{3}{2} + \frac{3 \sqrt{3} i}{2}$$
x3 = -3/2 + 3*sqrt(3)*i/2
Sum and product of roots
[src]
sum
___ ___
3 3*I*\/ 3 3 3*I*\/ 3
3 + - - - --------- + - - + ---------
2 2 2 2
$$\left(3 + \left(- \frac{3}{2} - \frac{3 \sqrt{3} i}{2}\right)\right) + \left(- \frac{3}{2} + \frac{3 \sqrt{3} i}{2}\right)$$
=
0
$$0$$
product
/ ___\ / ___\ | 3 3*I*\/ 3 | | 3 3*I*\/ 3 | 3*|- - - ---------|*|- - + ---------| \ 2 2 / \ 2 2 /
$$3 \left(- \frac{3}{2} - \frac{3 \sqrt{3} i}{2}\right) \left(- \frac{3}{2} + \frac{3 \sqrt{3} i}{2}\right)$$
=
27
$$27$$
27
Numerical answer
[src]
x1 = -1.5 + 2.59807621135332*i
x2 = -1.5 - 2.59807621135332*i
x3 = 3.0
x3 = 3.0
The graph