Prove that $(2+ \sqrt5)^{\frac13} + (2- \sqrt5)^{\frac13}$ is an integer

Question

When checked in calculator it is 1. But how to prove it? Also it is not a normal addition like $x+ \frac1x$ which needs direct rationalization. So I just need something to proceed.

Answer 1

Since $$(1 - \sqrt{5})^3 = 16 - 8 \sqrt{5}$$ and similarly $$(1 + \sqrt{5})^3 = 16 + 8 \sqrt{5}$$ it follows that $$(2 + \sqrt{5})^{1/3} + (2 - \sqrt{5})^{1/3} = \left(\frac {16+8\sqrt{5}} 8 \right)^{1/3} +\left(\frac {16-8\sqrt{5}} 8 \right)^{1/3} = \frac{1 + \sqrt{5}}{2} + \frac{1 - \sqrt{5}}{2} = 1.$$

Answer 2

Use the general formula $x^3+y^3=(x+y)\left((x+y)^2-3xy\right)$.

Here $x=\sqrt[3]{2+\sqrt{5}}$ and $y=\sqrt[3]{2-\sqrt{5}}$.

Let $a=\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}$ and

$$b=\sqrt[3]{2+\sqrt{5}}\sqrt[3]{2-\sqrt{5}}=\sqrt[3]{2^2-(\sqrt{5})^2}=-1$$

$$(2+\sqrt{5})+(2-\sqrt{5})=4$$

$$=a\left(a^2-3b\right)=a^3+3a$$

$$a^3+3a-4=0$$

You can apply Rational Root Theorem/Test, polynomial division, fundamental theorem of algebra, etc.

$$(a-1)\left(a^2+a+4\right)=0$$

$a^2+a+4=0$ has no real solutions, but clearly $a$ is real, so $a=1$.

Answer 3

We have $x- \left(2+\sqrt 5 \right)^{\frac{1}{3}} - \left(2-\sqrt 5 \right)^{\frac{1}{3}} = 0$

So $x^3 -(2-\sqrt 5)-(2+\sqrt 5) = 3x \left(2+\sqrt 5 \right)^{\frac{1}{3}}\left(2-\sqrt 5 \right)^{\frac{1}{3}}= -3x$

(invoking that $a^3+b^3+c^3 = 3abc$ when $a+b+c = 0$)

from which we see that $x$ is a root of $x^3+3x-4 =0$

Since the derivative is positive, this means it has only one real root, which by inspection is $x=1$

Answer 4

If $x = a+b$, then $x^3 = (3ab)x + (a^3+b^3)$. (Conversely, if $x^3 - px - q = 0 $ then we may find $a,b$, such that $3ab = p, a^3+b^3=q$. using Cardano's method for solving the cubic equation.)

Letting $w = e^{\frac {2i \pi}{3}} $ denote a cube root of unity, notice that the numbers $a+b, wa + w^2b$, and $w^2a + wb$ all satisfy the same cubic equation, $x^3 = 3abx + (a^3+b^3)$, and in general the solutions to the equation $x^3 = 3w^iabx + (a^3+b^3) $ are given by $x = w^ja + w^{i-j}b$, for $j=0,1,2$.

There are thus $9$ possible values of $(2+\sqrt{5})^{1/3} + (2-\sqrt{5})^{1/3}$, depending on which cube roots you take, and they each satisfy one of $3$ cubic equations. But only one of these values, namely $1$, is real, let alone rational. Hope it helps.

Answer 5

An easy way is to simplify the nested radical, and then combine like terms.

An easy way to denest $\sqrt[m]{A+B\sqrt[n]{C}}$ is to assume the form $a+b\sqrt[n]{C}$ and expand both sides with the binomial theorem. $\sqrt[3]{2+\sqrt5}$ becomes$$2+\sqrt5=(a^3+15ab^2)+(3a^2b+5b^3)\sqrt5\tag1$$ From which we get a system of equations\begin{align*} & a^3+15ab^2=2\tag2\\ & 3a^2b+5b^3=1\tag3\end{align*} Cross multiplying, we get$$a^3+15ab^2=6a^2b+10b^3\\a^3-6a^2b+15ab^2-10b^3=0\tag4$$ Dividing the last equation of $(4)$ by $b^3$ and substituting $x=a/b$ gives the cubic$$x^3-6x^2+15x-10=0\implies x=1$$ Therefore, $a=b$ and plugging that back into the system, we get$$a^3+15ab^2=2\implies 16b^3=2\implies a=b=\dfrac 12$$ Hence, $\sqrt[3]{2+\sqrt5}=\dfrac 12+\dfrac {\sqrt5} 2$ and similarly, $\sqrt[3]{2-\sqrt5}=\dfrac 12-\dfrac {\sqrt5}2$.

Therefore, we have$$\sqrt[3]{2+\sqrt5}+\sqrt[3]{2-\sqrt5}=\dfrac 12+\dfrac 12+\dfrac {\sqrt5}2-\dfrac {\sqrt5}2=\boxed{1}\tag5$$

Answer 6

Take the cube of the expression and expand it. You get \begin{multline} \left(\sqrt[3]{2 + \sqrt{5}} + \sqrt[3]{2 - \sqrt{5}}\right)^3 \\= 2 + \sqrt{5} + 2 - \sqrt{5} + 3\sqrt[3]{(2 + \sqrt{5})(2 - \sqrt{5})}\left(\sqrt[3]{2 + \sqrt{5}} + \sqrt[3]{2 - \sqrt{5}}\right) \\ = 4 - 3 \left(\sqrt[3]{2 + \sqrt{5}} + \sqrt[3]{2 - \sqrt{5}}\right) \end{multline} Thus, $\sqrt[3]{2 + \sqrt{5}} + \sqrt[3]{2 - \sqrt{5}}$ solves $x^3 = 4 - 3x$. It's not too difficult to find that $1$ is a solution to this equation.

Answer 7

Let $x=(2+\sqrt 5)^{1/3}$ and $y=(2-\sqrt 5)^{1/3} .$

We have $5>x^3>4$ and $0>y^3>-1$ so $2>x>1$ and $0>y>-1,$ so $2>x+y>0.$ Therefore $$x+y\in \mathbb Z \iff x+y=1.$$ We have $xy= (x^3y^3)^{1/3}=(-1)^{1/3}=-1$ and $x^3+y^3=4.$ So we have $$(x+y)^3=x^3+y^3+3xy(x+y)=4-3(x+y).$$ So $x+y$ is a positive solution to the equation $z^3+3z=4.$ The function $f(z)=z^3+3z$ is strictly increasing for $z>0$ (As it is the sum of the strictly increasing functions $z^3$ and $3z$.)

So $f(z)=4$ has at most one positive solution, which, if it exists, is equal to $x+y.$ Therefore $$x+y\in \mathbb Z \iff 1^3+3\cdot 1=4.$$

For insight on why this kind of thing happens, see Cardano's Method for solving cubic equations.

Answer 8

Not a full solution but if you use the equality: $(x+y)^3 = x^3 + y^3 + 3x^2y + 3y^2x$, you can reach the equation $z^3 + z = 4$ where $z$ is the value of the above expression. This leads to $z = 1$ as a solution. I cubed your formula and simplified. I do not, however, know how to solve $z^3 + z = 4$. Cubing both sides is safe only if $z$ is positive.