![Find $S$ where $S=\sqrt[3] {5+2 \sqrt {13}}+\sqrt[3]{5-2 \sqrt {13}}$, why am I getting an imaginary number?](https://pop.com.sg/uploads/find-s-where-s-sqrt-3-5-2-sqrt-13-sqrt-3-5-2-sqrt-13-why-am-i-getting-an-imaginary-number_720.jpg)
$\begingroup$$\large S=\sqrt[3] {5+2 \sqrt {13}}+\sqrt[3]{5-2 \sqrt {13}}$
Multiplying by conjugate:
$\large S=\dfrac {-3}{\sqrt[3] {5+2 \sqrt {13}}-\sqrt[3]{5-2 \sqrt {13}}}$
From the original:
$\large S-2\sqrt[3]{5-2 \sqrt {13}} =\sqrt[3] {5+2 \sqrt {13}}-\sqrt[3]{5-2 \sqrt {13}}$
Substituting:
$\large S=\dfrac{-3}{S-2\sqrt[3]{5-2 \sqrt {13}}}$
This leads to a quadratic equation in $\large S$ which I checked in wolframalpha and I got imaginary solutions. Why does this happen? I am not looking for an answer telling me how to solve this problem, I just want to know why this is wrong. Thanks.
$\endgroup$75 Answers
$\begingroup$You are not right: the conjugate to $S$ is $$ \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. $$
$\endgroup$$\begingroup$You can find quickly $S$ if you note that $$5+2\sqrt{13}=\left(\frac{1+\sqrt{13}}{2}\right)^3$$ and similarly for $5-2\sqrt{13}$.
$\endgroup$4$\begingroup$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 $S = 1$.
$\endgroup$$\begingroup$Note the identity $$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.
$\endgroup$2$\begingroup$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 $S=1$.
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