Sum of cube roots of two conjugate quadratic integers makes an integer.

Question

Consider the following expression:

$$(20-\sqrt{392})^{1/3}+(20+\sqrt{392})^{1/3}.$$

This equals $4$, but how can I show this?

Note that I do not want to make use of the following line of reasoning: 4 is a solution to $x^3-6x-40=0$, that this cubic has exactly one solution, and that all solutions to the cubic $x^3+px +q = 0$ are given by $$(-q/2-((q/2)^2+(p/3)^3)^{0.5})^{1/3}+(-q/2+((q/2)^2+(p/3)^3)^{0.5})^{1/3}.$$

I have noted the method set out here, How can I show that this complicated expression with square and cube roots reduces to the value 7? , but implementing it does not work very well.

I attempted this and got more complicated expressions than I started off with! It doesn't result in a unique solution for $a$ and $b$.

I do think that this is the right approach, though, i.e. making use of the fact that if $\sqrt c$ is irrational and $a+b\sqrt c=d+e \sqrt f$ then $a=d$ and $b=e$. (How do you prove this fact, by the way?)

Answer 1

First note that $392=2^3\times7^2$ and so $\sqrt{392}=14\sqrt{2}$, and hence $$\sqrt[3]{20+\sqrt{392}}=\sqrt[3]{20+14\sqrt{2}}.$$ Next, in the hope of finding a simple expression for the cube root, we compute $$(20+14\sqrt{2})(20-14\sqrt{2})=20^2-2\times14^2=8,$$ which shows that $$\sqrt[3]{20+14\sqrt{2}}\sqrt[3]{20-14\sqrt{2}}=\sqrt[3]{8}=2.$$ This suggests that perhaps there exist integers $a$ and $b$ such that $$(a+b\sqrt{2})^3=20+14\sqrt{2},$$ and then certainly $a$ and $b$ should satisfy $a^2-2b^2=2$. An obvious choice is $a=2$ and $b=1$, and indeed $$(2+\sqrt{2})^3=2^3+3\times2^2\times\sqrt{2}+3\times2\times\sqrt{2}^2+\sqrt{2}^3=20+14\sqrt{2},$$ and entirely analogously $(2-\sqrt{2})^3=20-14\sqrt{2}$. It follows that $$\sqrt[3]{20+14\sqrt{2}}+\sqrt[3]{20-14\sqrt{2}}=(2+\sqrt{2})+(2-\sqrt{2})=4.$$

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Alternatively, we could directly try to find integers $a$ and $b$ such that $$(a+b\sqrt{2})^3=20+14\sqrt{2}.$$ This boils down to the two simultaneous cubic equations \begin{eqnarray*} 20&=&a^3+6ab^2&=&a(a^2+6b^2),\\ 14&=&3a^2b+2b^3&=&b(3a^2+2b^2), \end{eqnarray*} from which it quickly follows that $a$ and $b$ are positive divisors of $20$ and $14$, respectively, and it is not hard to check that $a=2$, $b=1$ is the only solution.

Answer 2

kludginess and meta-cheating say hello.

alternative approach.

Let $~a ~\equiv ~\left[20 - \sqrt{392}\right]^{(1/3)}.$
Let $~b ~\equiv ~\left[20 + \sqrt{392}\right]^{(1/3)}.$
Let $~x ~\equiv a + b.$

The problem is to solve for $x$.

It is immediate that $~(ab) = (400 - 392)^{(1/3)} = 2.$

Therefore,

$$x^3 = (a+b)^3 = a^3 + b^3 + 3ab(x) = 40 + 6x.$$

Examining $~f(x) = x^3 - 6x - 40~$ gives the following:

• Clearly $~a > 0~$ and $~b > 0.~$ Therefore $x > 0.$

• $f(0) = -40, ~f(\sqrt{2}) < 0~,$ and by the OP's own observation, $f(4) = 0.$

• $f'(x) = 3x^2 - 6 = 3(x^2 - 2).~$ This means that in the interval $~[0, \sqrt{2}]~, f(x)~$ is strictly decreasing and in the interval $~[\sqrt{2}, +\infty), f(x)~$ is strictly increasing.

• Therefore, $~f(x)~$ can have at most 1 positive real root.

This proves that $x = 4.$