"Find the three cube roots of 27 and give the a answer in standard form."?

Question

can anybody tell me how to answer this problem? Is there an equation? really need the help, thanks.

..it says find the THREE cube roots of 27.. as in only one of them is 3.. there are still 2 others.

Answer 1

The equation to start with is : x^3 = 27
or, x^3 = 3^3
or, x^3 - 3^3 = 0

This is a difference of two cubes,
so it can be factored :

(x - 3)(x^2 + 3x + 9) = 0

One solution is seen to be x = 3.

The other solutions are obtained by
applying the quadratic formula to :

x^2 + 3x + 9 = 0

This gives :

x = {-3 ± sqrt[3^2 - 4(1)(9)]} / (2 * 1)

= [-3 ± sqrt(-27)] / 2

= [-3 ± i*3*sqrt(3)] / 2

= (-3 / 2) ± [3*sqrt(3) / 2] * i

Answer 2

3

Answer 3

Cube roots of 27 are 3, 3ω, 3ω² where ω, ω² are the two comlex cube roots of 1
ω = (-1 + i √3}/2 and ω² = (-1 - i √3}/2

These are the solutions to the equation
x² + x + 1 = 0 (as x³ = 1 means x³ - 1 = (x - 1)(x² + x + 1) = 0

Note: Some properties of ω and ω²
1. Since w³ = 1 = ω . ω² → ω² = 1/ω
2. Since ω is a solution to x² + x + 1 = 0 then 1 + ω + w³ = 0
3. 1, ω and ω² lie on a unit circle on an Argand diagram at angles of 0, 2π/3 and 4π/3

Answer 4

x^3 = 27
x^3 - 27 = 0
(x-3)(x^2 + 3x + 9) = 0 (Difference of 2 cubes formula)
(x-3) = 0 and (x^2 + 3x + 9) = 0 (Zero property)
x=3 and x = [-3 +/- root (9 - 36)] /2

So 1 real root, and 2 imaginary roots...
So similar to james k answer, but in a different form...

Answer 5

An alternative method for finding the complex roots is to exploit the way that multiplying complex numbers together adds the angle while multiplying the magnitude. You want to start off with something that is magnitude 3 and get back to the +ve real axis after three multiplications so it will be 3cos(2*pi/3)+i3sin(2*pi/3) or 3cos(-2pi/3)+i3sin(-2pi/3)

best of luck.

Answer 6

3x3=9x3=27, I guess standard form would be 3x3x3 = 27 or 3 to the 3rd(as an exponent) = 27. Good Luck!

Answer 7

27 to the third....27/3 =9.....9/3 = 3 check 3 x 3 = 9....9 x 3 = 27

Answer 8

3, 9 and 27 itself.

Answer 9

a cube root is numberXnumberXnumber soooo.......start with some low numbers and see where you get....

2x2x2 = (2x2 = 4)x2 = 8...no
3x3x3 = (3x3 = 9)x3 = 27....woohoo

It's all about trial and error

Answer 10

³√27 = 3

Because

3 x 3 x 3 = 27

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