Question

Question #4a512

Answer 1

See the explanation below.

Explanation:

In your comment , this is what you are puzzled about:

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

For a reason not mentioned (perhaps relevant to the larger question where this arose) this has been factored using the following:

`a^3-b^3 = (a-b)(a^2+ab+b^2)`

`1=1^3` (`1` is a cube)

`sqrtx = x^(1/2) = x^(3/6) = (x^(1/6))^3` `" "` (we can write `sqrtx` as a cube)

`(x^(1/6))^2 = x^(1/3)`

And, finally, the terms of the trinomial factor have been re-ordered by commutativity of addition.

Note

For a different problem we might choose to factor using:

`sqrtx - 1 = (x^(1/4))^2 - 1^2`
and the difference of squares.

Note 2

We could, instead, think of `sqrtx` as `(x^(1/8))^4` and use:

`a^4-b^4 = (a-b)(a+b)(a^2+b^2)` to factor.

`sqrtx-1 = (x^(1/8)-1)(x^(1/8)+1)(x^(1/4)+1)`

`= (root8x-1)(root8x+1)(root4x+1)`

Note 3
For positive integer `n`,

`a^n-b^n = (a-b)(a^(n-1)+a^(n-2)b+ a^(n-3)b^2 + * * * +a^2b^(n-3)+ ab^(n-2)+b^(n-1))`