Find $S$ where $S=\sqrt[3] {5+2 \sqrt {13}}+\sqrt[3]{5-2 \sqrt {13}}$, why am I getting an imaginary number?

Question

$\large S=\sqrt[3] {5+2 \sqrt {13}}+\sqrt[3]{5-2 \sqrt {13}}$

Multiplying by conjugate:

$\large S=\dfrac {-3}{\sqrt[3] {5+2 \sqrt {13}}-\sqrt[3]{5-2 \sqrt {13}}}$

From the original:

$\large S-2\sqrt[3]{5-2 \sqrt {13}} =\sqrt[3] {5+2 \sqrt {13}}-\sqrt[3]{5-2 \sqrt {13}}$

Substituting:

$\large S=\dfrac{-3}{S-2\sqrt[3]{5-2 \sqrt {13}}}$

This leads to a quadratic equation in $\large S$ which I checked in wolframalpha and I got imaginary solutions. Why does this happen? I am not looking for an answer telling me how to solve this problem, I just want to know why this is wrong. Thanks.

What do you mean why, you can multiply the expression by whatever you want. — Ovi, Jun 28 '13 at 14:28
It looks to me like you may have taken the conjugate in the square-root sense. But these are cube roots. — Cheerful Parsnip, Jun 28 '13 at 14:29
@GrumpyParsnip Yes thanks I just figured that out, the -3 in the numerator is wrong — Ovi, Jun 28 '13 at 14:29
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score 8 · Accepted Answer · answered Jun 28 '13 at 14:29

8

You are not right: the conjugate to $S$ is $$ \left[\sqrt[3] {5+2 \sqrt {13}}\right]^2-\sqrt[3]{5+2 \sqrt {13}}\sqrt[3] {5-2 \sqrt {13}}+\left[\sqrt[3]{5-2 \sqrt {13}}\right]^2. $$

answered Jun 28 '13 at 14:29

Boris Novikov

17,470

score 5 · Answer 2 · answered Jun 28 '13 at 15:02

5

You can find quickly $S$ if you note that $$5+2\sqrt{13}=\left(\frac{1+\sqrt{13}}{2}\right)^3$$ and similarly for $5-2\sqrt{13}$.

answered Jun 28 '13 at 15:02

Cortizol

3,669

wow${}{}{}{}{}{}$ – Lord Soth Jun 28 '13 at 15:11
How did you get that? – JSCB Jun 29 '13 at 02:19
@ᴊᴀsᴏɴ The standard trick that I saw in high school. – Cortizol Jun 29 '13 at 09:00
You assume that $a,b$ satisfy $5+2\sqrt{13} = (a+b\sqrt{13})^3 = (a^3+39ab^2)+(3a^2b+13b^3)\sqrt{13}$ and try to find nice values for $a$ and $b$. As it turns out, $a = b = \frac{1}{2}$ work. – JimmyK4542 Nov 09 '14 at 00:54

score 4 · Answer 3 · answered Jun 28 '13 at 14:44

It's probably better to cube $S$: $$S^3 = (5 + 2\sqrt{13}) + 3\sqrt[3] {5+2 \sqrt {13}}\sqrt[3] {5- 2 \sqrt {13}}\bigg(\sqrt[3] {5+2 \sqrt {13}} + \sqrt[3] {5-2 \sqrt {13}}\bigg) + (5 - 2\sqrt{13})$$ $$= 10 + 3\sqrt[3] {5^2-(2 \sqrt {13})^2}S$$ $$= 10 -9S$$ So $S$ satisfies the equation $S^3 + 9S -10 = 0$. The polynomial $S^3 + 9S -10$ factorizes as $(S-1)(S^2 + S - 10) = 0$, so you expect that $S = 1$.

But you still have to make sure $S$ is not one of the roots of $S^2 + S - 10 = 0$, given by ${\displaystyle -{1 \over 2} \pm {\sqrt{41} \over 2}}$. For this, note that the absolute value of each of these roots is greater than ${\sqrt{41} \over 2} - {1 \over 2} > {\sqrt{36} \over 2} - {1 \over 2} > 2$. On the other hand $\sqrt[3] {5+2 \sqrt {13}}$ is less than $\sqrt[3] {5+2 \sqrt {16}} = \sqrt[3]{13} < 2$. The other term $\sqrt[3] {5-2 \sqrt {13}}$ is negative and of smaller absolute value, so we conclude that $0 < S < 2$, and the only possibility is that $S = 1$.

score 2 · Answer 4 · answered Jun 28 '13 at 14:52

2

Note the identity $$x^3+y^3+z^3=(x+y+z)(x^2+y^2+z^2-xy-yz-xz)+3xyz$$

So that if $x+y+z=0$ we have $x^3+y^3+z^3=3xyz$

This is often useful with cubic expressions like this. Take $x=-S;\text{ } y=\sqrt[3] {5+2 \sqrt {13}};\text{ }z=\sqrt[3]{5-2 \sqrt {13}}$

Then we have$$-S^3+5+2\sqrt{13}+5-2\sqrt{13}=-3S\sqrt[3]{25-52}$$ whence $$S^3+9S-10=0$$

and you recover the cubic of which $S$ is a root. As it happens this is easy to factor, and you need to identify the root which corresponds to the expression you were given. I shall not repeat what Zarrax has put from this point on. But I think the trick is worth knowing, as it hugely simplifies the arithmetic.

answered Jun 28 '13 at 14:52

Mark Bennet

100,194

Thanks, but how do you factor a cubic equation such as this? Do you test for possible rational roots? – Ovi Jun 28 '13 at 15:00
@Ovi - indeed, as Zarrax has shown. When these expressions are set as puzzles they normally turn out well. The original expression tends to be what you get by plugging the coefficients into the formula for the roots of a (reduced - ie no $x^2$ term) cubic - and this method recovers the cubic in question. The only real reason for asking the question is that there is an easier way of finding the root than using the formula. Examining the method here will illuminate the connection. – Mark Bennet Jun 28 '13 at 15:17

score 1 · Answer 5 · answered Nov 09 '14 at 00:39

For simplicity, we'll let $S = a+b$ where $a=\sqrt[3]{5+2\sqrt{13}}$ and $b=\sqrt[3]{5-2\sqrt{13}}$.

Note that $S^3=a^3+3a^2b+3ab^2+b^3=5+2\sqrt{13}+5-2\sqrt{13}+3a^2b+3ab^2$.

$S^3+9S-10=3a^2b+3ab^2+9a+9b=3a(ab+3)+3b(ab+3)=3(a+b)(ab+3)=3S(ab+3)$.

Now, $ab=\sqrt[3]{(5+2\sqrt{13})(5-2\sqrt{13})}=\sqrt[3]{25-52}=\sqrt[3]{-27}=-3$. So $ab+3=0$. Thus, we see that $S$ is a root of the cubic equation $x^3+9x-10$. It is clear that $1$ is a root so the cubic is equal to $(x-1)(x^2+x+10)$. By the Quadratic Formula, the other two roots are imaginary. Thus $S=1$.

Find $S$ where $S=\sqrt[3] {5+2 \sqrt {13}}+\sqrt[3]{5-2 \sqrt {13}}$, why am I getting an imaginary number?

5 Answers5