5
$\begingroup$

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.

$\endgroup$
4
$\begingroup$

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 $3$.

$\endgroup$
  • $\begingroup$ Thats very nice! $\endgroup$ – Rudi_Birnbaum Sep 23 '17 at 16:59
  • $\begingroup$ @R_Berger My reaction when I first learnt this trick. $\endgroup$ – user8277998 Sep 23 '17 at 17:01
  • $\begingroup$ That gives some idea on field extensions, and contains in deed potential to generalization, I think. $\endgroup$ – Rudi_Birnbaum Sep 23 '17 at 17:48
  • $\begingroup$ @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. $\endgroup$ – user8277998 Sep 23 '17 at 17:56
  • $\begingroup$ @MichaelRozenberg No problem. Cheers :). $\endgroup$ – user8277998 Sep 23 '17 at 18:10
5
$\begingroup$

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 $$a^3+3a-36=(a-3)(a^2+3a+12)=0 \to a=3$$

$\endgroup$
  • $\begingroup$ If you remove the dot (its less than 6 char edit) then its the answer. $\endgroup$ – Rudi_Birnbaum Sep 23 '17 at 16:55
3
$\begingroup$

$$\left(\sqrt[3]{18+5\sqrt{13}} + \sqrt[3]{18-5\sqrt{13}}\right)^3=$$ $$=18+5\sqrt{13}+18-5\sqrt{13}+3\sqrt[3]{18+5\sqrt{13}} \sqrt[3]{18-5\sqrt{13}}\left(\sqrt[3]{18+5\sqrt{13}} + \sqrt[3]{18-5\sqrt{13}}\right)=$$ $$=36+3\sqrt[3]{-1}\left(\sqrt[3]{18+5\sqrt{13}} + \sqrt[3]{18-5\sqrt{13}}\right)=$$ $$=36-3\left(\sqrt[3]{18+5\sqrt{13}} + \sqrt[3]{18-5\sqrt{13}}\right).$$ 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!

$\endgroup$
  • $\begingroup$ You mean cubing? $\endgroup$ – Kenny Lau Sep 23 '17 at 16:44
  • $\begingroup$ Thanks guys! I fixed. $\endgroup$ – Michael Rozenberg Sep 23 '17 at 17:01
2
$\begingroup$

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.

$\endgroup$
  • 2
    $\begingroup$ In the square case, I would have taken the danger on me. $\endgroup$ – Rudi_Birnbaum Sep 23 '17 at 16:50
  • $\begingroup$ Seems like we have a real badass here :P $\endgroup$ – user370967 Sep 23 '17 at 20:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.