Question

How do you factor `x^3-1/x^3+4` ?

Answer 1

`x^3-1/x^3+4 = (x-1/x+1)(x^2-x+2+1/x+1/x^2)`

Explanation:

Given:

`x^3-1/x^3+4`

Here's a partial factorisation...

Start with:

`(x-1/x)^3 = x^3-3x+3/x-1/x^3`

So:

`x^3-1/x^3+4 = (x-1/x)^3+3(x-1/x)+4`

Let:

`t = x-1/x`

Then:

`t^3+3t+4 = (t+1)(t^2-t+4)`

So:

`x^3-1/x^3+4 = (x-1/x+1)((x-1/x)^2-(x-1/x)+4)`

`color(white)(x^3-1/x^3+4) = (x-1/x+1)(x^2-2+1/x^2-x+1/x+4)`

`color(white)(x^3-1/x^3+4) = (x-1/x+1)(x^2-x+2+1/x+1/x^2)`

Answer 2

`x^3-1/x^3+4`

`=1/x^3(x+(1/2-1/2sqrt(5)))(x^2-(1/2-1/2sqrt(5))x+(3/2-1/2sqrt(5)))(x+(1/2+1/2sqrt(5)))(x^2-(1/2+1/2sqrt(5))x+(3/2+1/2sqrt(5)))`

Explanation:

Given:

`x^3-1/x^3+4 = 1/x^3(x^6+4x^3-1)`

Let us avoid the rational expressions while we factor, by focusing on the hextic polynomial `x^6+4x^3-1`.

We can first treat this as a quadratic in `x^3` and factor by completing the square:

`x^6+4x^3-1 = (x^3)^2+4(x^3)+4-5`

`color(white)(x^6+4x^3-1) = (x^3+2)^2-(sqrt(5))^2`

`color(white)(x^6+4x^3-1) = (x^3+2-sqrt(5))(x^3+2+sqrt(5))`

Note that:

`x^3+y^3 = (x+y)(x^2-xy+y^2)`

Using this, we can factor further in terms of `root(3)(2-sqrt(5))` and `root(3)(2+sqrt(5))`, but these expressions are overcomplicated.

Note that:

`(1+sqrt(5))^3 = 1+3sqrt(5)+3(sqrt(5))^2+(sqrt(5))^3 = 16+8sqrt(5) = 2^3(2+sqrt(5))`

Similarly:

`(1-sqrt(5))^3 = 2^3(2-sqrt(5))`

So:

`root(3)(2+sqrt(5)) = 1/2+1/2sqrt(5)`

`root(3)(2-sqrt(5)) = 1/2-1/2sqrt(5)`

Then:

`(1/2+1/2sqrt(5))^2 = 1/4(1+sqrt(5))^2 = 1/4(1+2sqrt(5)+5) = 3/2+1/2sqrt(5)`

`(1/2-1/2sqrt(5))^2 = 1/4(1-sqrt(5))^2 = 1/4(1-2sqrt(5)+5) = 3/2-1/2sqrt(5)`

So putting `y = 1/2-1/2sqrt(5)` or `y = 1/2+1/2sqrt(5)` we find:

`(x^3+2-sqrt(5)) = (x+(1/2-1/2sqrt(5)))(x^2-(1/2-1/2sqrt(5))x+(3/2-1/2sqrt(5)))`

`(x^3+2+sqrt(5)) = (x+(1/2+1/2sqrt(5)))(x^2-(1/2+1/2sqrt(5))x+(3/2+1/2sqrt(5)))`

Putting it all together:

`x^3-1/x^3+4`

`=1/x^3(x+(1/2-1/2sqrt(5)))(x^2-(1/2-1/2sqrt(5))x+(3/2-1/2sqrt(5)))(x+(1/2+1/2sqrt(5)))(x^2-(1/2+1/2sqrt(5))x+(3/2+1/2sqrt(5)))`