How does $\sqrt[6]{26+15\sqrt3}-\sqrt[6]{26-15\sqrt3}$ become $\sqrt2$?

Question

How can I simplify this?

$$\sqrt[6]{26+15\sqrt3}-\sqrt[6]{26-15\sqrt3}$$

I know the answer is $\sqrt2$, but don't know where to start.

Thx.

Answer 1

Rewrite the expressions under radicals as follows $$26\pm15\sqrt3=\frac{1}{8}(\sqrt3\pm1)^6,$$ then $$\sqrt[6]{26+15\sqrt3}-\sqrt[6]{26-15\sqrt3}=\frac{1}{\sqrt[6]8}(\sqrt3+1)-\frac{1}{\sqrt[6]8}(\sqrt3-1)=\sqrt2$$

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EDIT

In school exercises, sometimes it is useful to find a different writing of the given expression in order to get a nice result. Here we know that the result is nice and simple.

First of all, realize that $\sqrt[6]A=\sqrt[3]{\sqrt A}.$
It would be nice to find rational numbers $p,q$ so that $$26\pm15\sqrt3=(p\pm q\sqrt3)^2.$$ However, $(a+b\sqrt3)^2=a^2+3b^2+2ab\sqrt3,$ and I was hoping to find integers $a,b.$ Thus I multiplied LHS by $2$ and solved $$52+30\sqrt3=a^2+3b^2+2ab\sqrt3 \quad (a,b\in\mathbb{Q}),$$ that is equivalent to the system \begin{cases} a^2+3b^2&=&52 \quad &\text{rational part}\\ 2ab\sqrt3&=&30\sqrt3 \quad &\text{irrational part} \end{cases} That gives $$2(26+15\sqrt3)=(5+3\sqrt3)^2$$ For $(26-15\sqrt3)$ we get $a,b$ with opposite signs: $$2(26-15\sqrt3)=(5-3\sqrt3)^2$$ Sixth root is positive, hence $$\sqrt[6]{26\pm15\sqrt3}=\frac{1}{\sqrt[6]2}\sqrt[3]{3\sqrt3\pm5}.$$

Again, I tried to find integers $x,y$ such that $$(x+y\sqrt3)^3=k(3\sqrt3+5)$$ for a constant $k,$ preferably integral. The principle of computation was as above. The constant turned out to be again $k=2.$

Answer 2

Here is a useful technique using the following

If $a+b+c=0$ then $a^3+b^3+c^3=3abc \tag{1}$

Now, let's note $$x=\sqrt[6]{26+15\sqrt3}-\sqrt[6]{26-15\sqrt3}$$ Obviously (or easy to check) $$x>0 \tag{2}$$ Then, using $\left(26+15\sqrt{3}\right)\left(26-15\sqrt{3}\right)=1$, we have $$x^2+2=\sqrt[3]{26+15\sqrt3}+\sqrt[3]{26-15\sqrt3}$$ or $$(x^2+2)+\left(-\sqrt[3]{26+15\sqrt3}\right)+\left(-\sqrt[3]{26-15\sqrt3}\right)=0$$ From $(1)$ this leads to $$(x^2+2)^3-52=3(x^2+2)\tag{3}$$ It's not too difficult to check that $-\sqrt{2}$ and $\sqrt{2}$ are roots of $(3)$.

Now, we divide $(3)$ (polynomial division) by $x^2-2$ and obtain $$(2) \iff (x^2 - 2) (x^4 + 8 x^2 + 25) = 0$$ Or $\sqrt{2}$ is the only positive root, because $x^4 + 8 x^2 + 25 > 0$ and has no real roots. Thus, from $(2)$ we conclude that $x=\sqrt{2}$.

Answer 3

First thing, notice $\sqrt[6]{26+15\sqrt3}$ and $\sqrt[6]{26-15\sqrt3}$ are reciprocals of each other. (Just multiply them together.) So if you let $a = \sqrt[6]{26+15\sqrt3}$, you are seeking to show that $a - {1 \over a} = \sqrt{2}$.

Squaring both sides, given that your expression is positive you can rewrite what you are trying to show as $$a^2 + \frac{1}{a^2} = 4$$ Let $x = a^2 + {1 \over a^2}$. Then $$x^3 = a^6 + {1 \over a^6} + 3\bigg(a^2 + {1 \over a^2}\bigg)$$ Note that $a^6 + {1 \over a^6} = 26+15\sqrt3 + 26-15\sqrt3 = 52$, so the above gives $$x^3 = 52 + 3x$$ The roots of $x^3 - 3x - 52 = 0$ are $4$, $-2 + 3i$, and $-2 - 3i$. Since $x$ is real, we must have $x = 4$ as needed.

Answer 4

Let $A=\sqrt[6]{26+15\sqrt3}-\sqrt[6]{26-15\sqrt3}$. Then $A^2=B-2$ where $B=\sqrt[3]{26+15\sqrt3}+\sqrt[3]{26-15\sqrt3}$.

On the other hand, $B$ satisfies the equation $B^3-3B-52=0$. We see that it has one real root which is $B=4$. Hence, $A=\sqrt2$.