Mister Exam
Other calculators
-
How to use it?
- Equation:
- Equation x-y=1
-
Equation (x+2)/9=(x-3)/2
-
Equation x^2-4x+5=0
-
Equation x^3=27
- Express {x} in terms of y where:
- 18*x-8*y=-2
- 4*x+2*y=5
- -17*x+1*y=-11
- 20*x+7*y=-9
- Derivative of:
-
x^3
- Integral of d{x}:
-
x^3
- x^3
- Sum of series:
-
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
- 0,2(7-2y)=2,3—0,3(y—6)
- 16x^3-x=0
- sqrt((150/x^2+252/x^3)^2+4*(135/(0,208*x^3))^2)=160
- 0,2(7-2y)=2,3-0,3(y-6)
- (x^3+4)/x^2=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