Calculate the radicals (order 3)

Question

Calculate:

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

I tried to note that expression with x and then cube it but then it becomes a complicated calculation

Answer 1

Hint: In general,

$$(a+b)^3=a^3+3a^2b+3ab^2+b^3$$

Now, in this specific case, we may write it as $(a+b)^3=a^3+3ab(a+b)+b^3$.

Take $a=\sqrt[3]{5+2\sqrt{13}}$, $b=\sqrt[3]{5-2\sqrt{13}}$, then we get

\begin{align*}ab &= \sqrt[3]{5+2\sqrt{13}}\sqrt[3]{5-2\sqrt{13}} \\ &=\sqrt[3]{(5+2\sqrt{13})(5-2\sqrt{13})} \\ &=\sqrt[3]{5^2-(2\sqrt{13})^2}\\ &=\sqrt[3]{-27}=-3 \end{align*}

Take $x=a+b$. We want to know $x$. We get the equation

$$x^3=5+2\sqrt{13}-9x+5-2\sqrt{13}$$

$$x^3=-9x+10$$

Now try to find a root of this equation.

Answer 2

As I suspected might happen, someone -- wythagoras -- managed to get an answer posted in the time I spent off-line writing this up. However, maybe the less clever algebraic drudgery method I used will be of help to others.

Begin by putting $\;x = \sqrt[3]{5+2 \sqrt{13}} + \sqrt[3]{5-2 \sqrt{13}}.$ Now cube both sides:

$$x^3 \;\; = \;\; \left(\sqrt[3]{5+2 \sqrt{13}} + \sqrt[3]{5-2 \sqrt{13}}\right)^3$$

$$x^3 = \left(\sqrt[3]{5+2 \sqrt{13}}\right)^3 + 3\cdot \left(\sqrt[3]{5+2 \sqrt{13}}\right)^2 \sqrt[3]{5-2 \sqrt{13}} + 3\cdot \sqrt[3]{5+2 \sqrt{13}} \left(\sqrt[3]{5-2 \sqrt{13}}\right)^2 + \left(\sqrt[3]{5-2 \sqrt{13}}\right)^3$$

In the above I made use of $\;(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^{3}.$ Rewriting the right-hand side a little gives

$$x^3 \;\; = \;\; 5 + 2 \sqrt{13} \;\; + \;\; 3 \sqrt[3]{(5+2 \sqrt{13})(5+2 \sqrt{13})(5-2 \sqrt{13})} + 3 \sqrt[3]{(5+2 \sqrt{13})(5-2 \sqrt{13})(5-2 \sqrt{13})} \;\; + \;\; 5 - 2 \sqrt{13}$$

Now use the fact that $\;(5+2 \sqrt{13})(5-2 \sqrt{13}) \; = \; 25- 4\cdot 13 = -27\;$ to get

$$x^3 \;\; = \;\; 10 \;\; + \;\; 3 \sqrt[3]{(5+2 \sqrt{13})(-27)} \; + \; 3 \sqrt[3]{ (-27) (5-2 \sqrt{13})}$$

$$x^3 \;\; = \;\; 10 \;\; - \;\; 9 \sqrt[3]{5+2 \sqrt{13}} \; - \; 9 \sqrt[3]{ 5-2 \sqrt{13}}$$

$$x^3 \;\; = \;\; 10 \;\; - \;\; 9 \left(\sqrt[3]{5+2 \sqrt{13}} \; + \; \sqrt[3]{ 5-2 \sqrt{13}}\right)$$

$$x^3 + 9x - 10 \; = \; 0$$

By lucky recognition, or by a systematic use of the rational root test, we see that $x=1$ is a solution. Therefore, by the Factor Theorem, we know that $x-1$ is a factor of the left-hand side. Using high school algebra long division (or use synthetic division if you remember how to do it) to divide $x-1$ into $ x^3 + 9x - 10,$ we get

$$(x-1)(x^2 + x + 10) \; = \; 0$$

Technical Note: If we assume that the original expression represents a real number, then that real number is $1.$ However, because each real number besides $0$ has three different cube roots when working with complex numbers, the original expression could conceivably represent $9$ different complex numbers. (Before adding, there are $9$ ways to simultaneously pick one of the $3$ cube roots of $5+2 \sqrt{13}$ and one of the $3$ cube roots of $5-2 \sqrt{13}.)$ However, I believe it can be shown that several of these choices of picking cube roots lead to the same complex number sum, and the original expression represents exactly $3$ numbers, one being $x=1$ and the other two being the two solutions to $x^2 + x + 10 = 0.$