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The following is a question that I composed and solved.
I want to know if it is mathematically correct and well-formatted.

Question

Show that $\root{3}\of{\sqrt{5}+2}+\root{3}\of{\sqrt{5}-2}=\sqrt{5}$

Answer

Guess that: $\root{3}\of{\sqrt{5}\pm 2}=a\pm b$

Then $\root{3}\of{\sqrt{5}+2}+\root{3}\of{\sqrt{5}-2}=$
$(a+b)+(a-b)=2a=\sqrt{5} \implies a=\frac{1}{2}\sqrt{5}$

Using $a=\frac{1}{2}\sqrt{5}$, we make a revised guess:

$\root{3}\of{\sqrt{5}\pm 2}=\frac{1}{2}\sqrt{5}\pm c$

Now solve for c:

$\begin{align} \left(\root{3}\of{\sqrt{5}\pm 2}\right)^3&=\left(\frac{1}{2}\sqrt{5}\pm c\right)^3 \\ \sqrt{5}\pm 2&=\left(\frac{3}{2}c^2+\frac{5}{8}\right)\sqrt{5}\pm\left(c^3+\frac{15}{4}c\right) \\ \end{align}$

Equating the irrational and rational parts,
we get a system of 2 equations:

(1) $\quad \frac{3}{2}c^2+\frac{5}{8}=1 \implies c=\pm \frac{1}{2}$
(2) $\quad c^3+\frac{15}{4}c=2 \implies$ Only $c=\frac{1}{2}$ works

So $\root{3}\of{\sqrt{5}\pm 2}=\frac{1}{2}\sqrt{5}\pm \frac{1}{2}$

Finally, $\root{3}\of{\sqrt{5}+2}+\root{3}\of{\sqrt{5}-2}=$
$\left(\frac{1}{2}\sqrt{5}+\frac{1}{2}\right)+\left(\frac{1}{2}\sqrt{5}-\frac{1}{2}\right)=\sqrt{5}$

Q. E. D.

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    $\begingroup$ Since you're saying "guess that...", then everything is proved on the basis that your initial guess was correct. So, at the end of your train of thought, it is still left to prove that your guess is correct. This can be shown by picking $a$ as the midpoint between the two value and $b$ as half of their distance. Had you said something like "notice that" instead of "guess that" I would have had nothing to object. $\endgroup$
    – Bob
    Commented May 25, 2022 at 5:03
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    $\begingroup$ Your answer is correct. Good job ! $\endgroup$
    – Wang YeFei
    Commented May 25, 2022 at 5:34
  • $\begingroup$ It is correct! Have a look t this question for a more general method fo find a solution. $\endgroup$
    – Paul Frost
    Commented Jun 10, 2023 at 9:27

2 Answers 2

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Alternatively, you can proceed as follows: \begin{align*} x = \sqrt[3]{\sqrt{5} + 2} + \sqrt[3]{\sqrt{5} - 2} & \Longleftrightarrow x^{3} = 2\sqrt{5} + 3x\\\\ & \Longleftrightarrow x^{3} - 3x - 2\sqrt{5} = 0\\\\ & \Longleftrightarrow (x^{3} - 5x) + (2x - 2\sqrt{5}) = 0\\\\ & \Longleftrightarrow x(x + \sqrt{5})(x - \sqrt{5}) + 2(x - \sqrt{5}) = 0\\\\ & \Longleftrightarrow (x^{2} + x\sqrt{5} + 2)(x - \sqrt{5}) = 0\\\\ & \Longleftrightarrow \left[\left(x + \frac{\sqrt{5}}{2} \right)^{2} + \frac{3}{4}\right](x - \sqrt{5}) = 0\\\\ & \Longleftrightarrow x = \sqrt{5}. \end{align*}

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As an alternative, since $\sqrt{5}+2=\frac1{\sqrt{5}-2}$ we have

$$\sqrt[3]{\sqrt{5}+2}+\sqrt[3]{\sqrt{5}-2}=\sqrt 5\iff \sqrt[3]{\sqrt{5}+2}+\frac{1}{\sqrt[3]{\sqrt{5}+2}}=\sqrt 5$$

which leads to

$$\sqrt[3]{\sqrt{5}+2}=\frac{\sqrt 5 +1}2 \iff \left(\frac{\sqrt 5 +1}2\right)^3=\sqrt{5}+2$$

which is true.

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