# Why is $\sqrt{18+5\sqrt{13}} + \sqrt{18-5\sqrt{13}} = 3$?

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

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 $(\sqrt{18+5\sqrt{13}})^2 - (\sqrt{18-5\sqrt{13}})^2$, but no success. Then I have solved for one cubic root and put all to the third power. Also no success.

• The "trick" $(a+b)(a-b) = a^2-b^2$ that works for square roots will not work for cube roots. Maybe you want to think about the analogous formula $(a+b)(a^2-ab+b^2) = a^3+b^3$. Sep 23, 2017 at 16:45
• @TedShifrin you appear here but not in chat ;_; Sep 23, 2017 at 16:46
• The other two solutions to the equation are $-2\pm\sqrt{3}$, so this is perhaps some clever use of Vieta's formulas. Sep 23, 2017 at 16:51
• Sep 24, 2017 at 11:40
• The more recent question How to prove that $(-18+\sqrt{325})^{\frac{1}{3}}+(-18-\sqrt{325})^{\frac{1}{3}} = 3$ is very, very nearly a duplicate of this one. Mar 7, 2019 at 1:02

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

$$\sqrt{(18 + 5\sqrt{13})} = \dfrac 32 +\dfrac12\sqrt{13}$$

Similarly,

$$\sqrt{(18 - 5\sqrt{13})} = \dfrac 32 -\dfrac12\sqrt{13}$$

Hence the sum is $3$.

• Thats very nice! Sep 23, 2017 at 16:59
• @R_Berger My reaction when I first learnt this trick. Sep 23, 2017 at 17:01
• That gives some idea on field extensions, and contains in deed potential to generalization, I think. Sep 23, 2017 at 17:48
• @MichaelRozenberg I don't doubt that it is a bit ugly . I outlined the general method which OP can use elsewhere, for instance when he does not know the sum beforehand or when he had to denest radicals. No doubt your method is a bit easy here and I had +1 on your answer anyways. Sep 23, 2017 at 17:56
• @MichaelRozenberg No problem. Cheers :). Sep 23, 2017 at 18:10

Hint :$$a=\sqrt{18+5\sqrt{13}} + \sqrt{18-5\sqrt{13}}=b+c \\a^3=b^3+c^3+3bc(b+c)\\ a^3=18+5\sqrt{13}+18-5\sqrt{13}+3\sqrt{18+5\sqrt{13}}.\sqrt{18-5\sqrt{13}}(a)\\a^3=36+3\sqrt{324-325}a\\a^3=36-3a$$solve for a $$a^3+3a-36=(a-3)(a^2+3a+12)=0 \to a=3$$

• If you remove the dot (its less than 6 char edit) then its the answer. Sep 23, 2017 at 16:55

$$\left(\sqrt{18+5\sqrt{13}} + \sqrt{18-5\sqrt{13}}\right)^3=$$ $$=18+5\sqrt{13}+18-5\sqrt{13}+3\sqrt{18+5\sqrt{13}} \sqrt{18-5\sqrt{13}}\left(\sqrt{18+5\sqrt{13}} + \sqrt{18-5\sqrt{13}}\right)=$$ $$=36+3\sqrt{-1}\left(\sqrt{18+5\sqrt{13}} + \sqrt{18-5\sqrt{13}}\right)=$$ $$=36-3\left(\sqrt{18+5\sqrt{13}} + \sqrt{18-5\sqrt{13}}\right).$$ Now, let $\sqrt{18+5\sqrt{13}} + \sqrt{18-5\sqrt{13}}=x.$

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 $$18+5\sqrt{13}+18-5\sqrt{13}-27-3\cdot(-3)\sqrt{18+5\sqrt{13}}\cdot\sqrt{18-5\sqrt{13}}=0$$ or $$36-27-3(-3)(-1)=0,$$ which is true.

Done!

• You mean cubing? Sep 23, 2017 at 16:44
• Thanks guys! I fixed. Sep 23, 2017 at 17:01

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.

• In the square case, I would have taken the danger on me. Sep 23, 2017 at 16:50
• Seems like we have a real badass here :P
– user370967
Sep 23, 2017 at 20:18