Why is $\sqrt[3]{18+5\sqrt{13}} + \sqrt[3]{18-5\sqrt{13}} = 3$? 
How to show that $$\sqrt[3]{18+5\sqrt{13}} + \sqrt[3]{18-5\sqrt{13}} = 3?$$

This equality comes from solving $$t^3 - 15 t - 4 = 0$$ using Cardanos fomula and knowing the solution $t_1=4$.
I have attempted multiplying the whole thing with $(\sqrt[3]{18+5\sqrt{13}})^2 - (\sqrt[3]{18-5\sqrt{13}})^2$, but no success. Then I have solved for one cubic root and put all to the third power. Also no success.
 A: Hint :$$a=\sqrt[3]{18+5\sqrt{13}} + \sqrt[3]{18-5\sqrt{13}}=b+c \\a^3=b^3+c^3+3bc(b+c)\\
a^3=18+5\sqrt{13}+18-5\sqrt{13}+3\sqrt[3]{18+5\sqrt{13}}.\sqrt[3]{18-5\sqrt{13}}(a)\\a^3=36+3\sqrt[3]{324-325}a\\a^3=36-3a$$solve for a 
$$a^3+3a-36=(a-3)(a^2+3a+12)=0 \to a=3$$
A: Let $(a + b\sqrt{13})^3 = (18 + 5\sqrt{13})$ for $a, b \in \Bbb Q$
Expanding the LHS gives,
$$(a^3 + 39 ab^2 - 18 ) +\sqrt{13}(3a^2 b + 13 b^3 - 5) = 0$$,
From this we get,
$$\begin{cases}a^3 + 39 ab^2 - 18 = 0 \\ 3a^2 b + 13 b^3 - 5 = 0\end{cases}$$
Solving the system give $ a = \dfrac 32$ and $ b = \dfrac12$
Therefore 
$$\sqrt[3]{(18 + 5\sqrt{13})} = \dfrac 32 +\dfrac12\sqrt{13}$$
Similarly,
$$\sqrt[3]{(18 - 5\sqrt{13})} = \dfrac 32 -\dfrac12\sqrt{13}$$
Hence the sum is $3$.
A: $$\left(\sqrt[3]{18+5\sqrt{13}} + \sqrt[3]{18-5\sqrt{13}}\right)^3=$$
$$=18+5\sqrt{13}+18-5\sqrt{13}+3\sqrt[3]{18+5\sqrt{13}} \sqrt[3]{18-5\sqrt{13}}\left(\sqrt[3]{18+5\sqrt{13}} + \sqrt[3]{18-5\sqrt{13}}\right)=$$
$$=36+3\sqrt[3]{-1}\left(\sqrt[3]{18+5\sqrt{13}} + \sqrt[3]{18-5\sqrt{13}}\right)=$$
$$=36-3\left(\sqrt[3]{18+5\sqrt{13}} + \sqrt[3]{18-5\sqrt{13}}\right).$$
Now, let $\sqrt[3]{18+5\sqrt{13}} + \sqrt[3]{18-5\sqrt{13}}=x.$
Thus, $$x^3=36-3x$$ or
$$x^3-3x^2+3x^2-9x+12x-36=0$$ or
$$(x-3)(x^2+3x+12)=0,$$ which gives $x=3$.
But I think the best way in this formulation it's the way by using
$$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-ac-bc).$$
Now, we need to prove that
$$18+5\sqrt{13}+18-5\sqrt{13}-27-3\cdot(-3)\sqrt[3]{18+5\sqrt{13}}\cdot\sqrt[3]{18-5\sqrt{13}}=0$$ or
$$36-27-3(-3)(-1)=0,$$
which is true.
Done!
A: Because the function $\mathbb{R} \to \mathbb{R}: x \mapsto x^3$ is injective, it follows that 
$x = y \iff x^3 = y^3$ (more precicely: $\Leftarrow$ follows from injectivity)
So, you can cube both sides without any danger, calculate, and make the conclusion you are looking for.
