Summing cube roots in fractions

Question

I found this problem, and understand the solution, but do not understand why they made the first assumption. The problem:

The first line of the solution says that:

The cube root of $1$ plus the cube root of $2$ plus the cube root of $4$ is a factor of $2-1$.

Why are you meant to assume this to solve the problem?

Answer 1

Probably they are telling you a way to rationalize the denominator so you can do the sum.

$1=2-1=(\sqrt[3]{2}-\sqrt[3]{1})(\sqrt[3]{4}+\sqrt[3]{2}+\sqrt[3]{1})$

Similarly

$1=3-2=(\sqrt[3]{3}-\sqrt[3]{2})(\sqrt[3]{9}+\sqrt[3]{6}+\sqrt[3]{4})$

and

$1=4-3=(\sqrt[3]{4}-\sqrt[3]{3})(\sqrt[3]{16}+\sqrt[3]{12}+\sqrt[3]{9})$

so you can multiply the first fraction by $\frac{\sqrt[3]{2}-\sqrt[3]{1}}{\sqrt[3]{2}-\sqrt[3]{1}}$, the second one by $\frac{\sqrt[3]{3}-\sqrt[3]{2}}{\sqrt[3]{3}-\sqrt[3]{2}}$ and the third one by $\frac{\sqrt[3]{4}-\sqrt[3]{3}}{\sqrt[3]{4}-\sqrt[3]{3}}$ and you get $$\frac{\sqrt[3]{2}-\sqrt[3]{1}}{2-1}+\frac{\sqrt[3]{3}-\sqrt[3]{2}}{3-2}+\frac{\sqrt[3]{4}-\sqrt[3]{3}}{4-3}=\sqrt[3]{2}-1+\sqrt[3]{3}-\sqrt[3]{2}+\sqrt[3]{4}-\sqrt[3]{3}=\sqrt[3]{4}-1$$

Answer 2

Hint to explain the quote:

use $x^3-1=(x-1)(1+x+x^2)$ with $x=\sqrt[3]2.$

Answer 3

They are pointing to the fact (it's not an assumption) that $$ (x^2+xy+y^2)(x-y) = x^3-y^3$$ that is $$ \frac{1}{x^2+xy+y^2} = \frac{x-y}{x^3-y^3}$$ If you use that for $x=1$, $y=\sqrt[3]{2}$, you get $$ \frac{1}{1+\sqrt[3]{2}+\sqrt[3]{4}} = \frac{\sqrt[3]{2}-1}{2-1} = \sqrt[3]{2}-1$$ To get the other two fractions use $x=\sqrt[3]{2}$, $y=\sqrt[3]{3}$ and $x=\sqrt[3]{3}$, $y=\sqrt[3]{4}$.