Question

Find the cube roots of 8?

Answer 1

`root(3)(8)=2`

Explanation:

We have:

`root(3)(8)`

`=root(3)(2^3)`

`=2`

Answer 2

Well where is another way in resolving...

Explanation:

Cube root of `8` is simply;

`root(3)8`

Note: `root(3)a = a^(1/3)`

Therefore;

`8^(1/3)`

`2^(3(1/3))`, `(2^3 = 8)`

`2^(3/3)`

`2^1`

`2`

Hope this helps!

Answer 3

The real cube root of `8` is `2`

The other, non-real cube roots of `8` are `-1+-sqrt(3)i`

Explanation:

Since the question asks for the cube root`color(red)(s)` of `8`, let's take a deeper look.

The function `f(x) = x^3` is a strictly monotonically increasing function from `(-oo, oo)` onto `(-oo, oo)`

Its graph looks like this:
graph{x^3 [-10, 10, -11, 11]}

We can form the graph of `f^(-1)(x) = root(3)(x)` by reflecting about the line `y=x` (effectively swapping `x` and `y`) to get:

graph{(y^3-x) = 0 [-10, 10, -5, 5]}

Now `2^3 = 8`, so `2` is a cube root of `8`, and since `f(x) = x^3` is one to one, it is the only real cube root of `8`.

In other words it is the only real zero of the polynomial:

`x^3-8`

As a result, this polynomial has a factor `(x-2)` and we find:

`x^3-8 = (x-2)(x^2+2x+4)`

What about the zeros of `x^2+2x+4`?

They are cube roots of `8` too, but they are not real valued.

We can find them by completing the square and using `i^2=-1`:

`0 = x^2+2x+4`

`color(white)(0) = x^2+2x+1+3`

`color(white)(0) = (x+1)^2+(sqrt(3))^2`

`color(white)(0) = (x+1)^2-(sqrt(3)i)^2`

`color(white)(0) = ((x+1)-sqrt(3)i)((x+1)+sqrt(3)i)`

`color(white)(0) = (x+1-sqrt(3)i)(x+1+sqrt(3)i)`

So they are `x=-1+-sqrt(3)i`

In the complex plane, where the `x` coordinate represents the real part and the `y` coordinate the imaginary part, the numbers `2`, `-1+sqrt(3)i` and `-1-sqrt(3)i` are all on the circle of radius `2` and form the vertices of an equilateral triangle.

graph{(x^2+y^2-4)((x-2)^2+y^2-0.01)((x+1)^2+(y-sqrt(3))^2-0.01)((x+1)^2+(y+sqrt(3))^2-0.01) = 0 [-5, 5, -2.5, 2.5]}