Factoring an irrational number into a cube

Question

I have a math problem in which I need to calculate : $$\sqrt[3]{2 + \sqrt{5}} + \sqrt[3]{2 - \sqrt{5}}$$

In one of the previous problems in which there were steps to calcuate, they separated the numbers so they get a cube like $(a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$, where $a^3 + 3a^2b + 3ab^2 + b^3 = 2 + \sqrt{5}$ so that cube and cube root cancel each other.

I guessed many numbers and didn't get the right one. Also I was wondering if there was other way to solve this problem or easier way to complete cube without having to guess until you get the right numbers.

Does this answer your question? Calculate simple expression: $\sqrt[3]{2 + \sqrt{5}} + \sqrt[3]{2 - \sqrt{5}}$ - found using an Approach0 search. Note that this answer , and this other one, there both use a similar technique to what you're suggesting, & what I used in my now deleted answer. ... — John Omielan, Jan 26 '24 at 01:10
I don't know what kind of "guessing" you did. The number is handed to you as a sum of two numbers, that is, it's handed to you in the form $a+b$, now all you have to do is cube it. No guessing needed. — Gerry Myerson, Jan 26 '24 at 01:15
(cont.) In addition to the $2$ other questions that one was closed as a duplicate of, there are additional site duplicates, e.g., Sum of two irrational numbers. There are also several AoPS questions regarding this, e.g., Cube root, How to find cube root, etc. Last, but not least, welcome to Math SE. — John Omielan, Jan 26 '24 at 01:16

score 2 · Accepted Answer · answered Jan 26 '24 at 00:42

Let $\phi$ denote the golden ratio as the bigger root satisfying $\phi^2-\phi-1=0$. It is also denoted as $\phi=\frac{1+\sqrt5}2$. Multiplying by $\phi$ and adding one: $$\phi^2-\phi-1=0\implies\phi^3-\phi^2-\phi=0\implies\phi^3=\phi^2+\phi=2\phi+1$$Note that $2\phi+1=2+\sqrt5$, therefore the first radicand is $\phi$. The second radicand can be proved to be the other root of $x^2-x-1=0$, which is the conjugate of $\phi$ very similarly. The sum of these two is just $1$ because the sum of two roots of a quadratic equation is the same as its leading coefficient, which is the answer.

Factoring an irrational number into a cube

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