Well, I can't find the example on how to solve this.  If I multiply $$ \dfrac{2}{\sqrt[3]{9}+\sqrt[3]{15}+\sqrt[3]{25}} $$ with  $$ \dfrac{\sqrt[3]{9}-\sqrt[3]{15}+\sqrt[3]{25}}{\sqrt[3]{9}-\sqrt[3]{15}+\sqrt[3]{25}} $$ or similar, it just gets even more complicated, and I get 4 terms instead of 3 in the denominator, and more can't be better...  Can someone tell me some principle by which all kinds of expressions with 3 or more terms and with different roots could be rationalised? On a test, I haven't got much time to use appropriate formula, as I can for square difference and so on, I need a principle.
91740
Rationalize   $\dfrac{1}{\sqrt[3]{p^2}+\sqrt[3]{pq}+\sqrt[3]{q^2}}.$  How would I go about doing this without wading through lots of algebra?  Is there a trick similar to how you can multiply by $\dfrac{\sqrt a-\sqrt b}{\sqrt a-\sqrt b}$ with square roots?  Thanks in advance!
80740
Using 5000-digit precision in PARI/GP, I discovered that the fractional part of $(\sqrt{2}+\sqrt{3})^{2008}$ is extremely small, less than $10^{-999}$.  Is there a simple explanation for this fact ?  This looks like a Pisot number issue (similar questions have already been studied on MSE, see for example ), but it’s a more complicated situation.   Related :
33388
I would really appreciate an explanation of the steps needed to differentiate expressions such as:  $$\sqrt{\frac{1-2x}{1+2x}}$$  $$\frac{2x}{\sqrt{x^2+7}}$$  $$\frac{5}{2+\sqrt{1+3x}}$$  I've been stuck on questions like these for a while and I can't seem to get my head around them.
113034
How to show that a root of the equation $$x (x+1)(x+2) ....... (x+2009) = c $$ can have multiplicity at most 2 , and to find the value of $ c $   for which this is possible.  I proceeded by using the derivative method, but the number of terms involved is huge.I have been trying to find some common property for all these terms, but I've come up with nothing so far.
77139
Motivated by the positive answer to the following question:      I'm curious about whether ${\Bbb Z}[\sqrt{2}+\sqrt{3}]=\Bbb{Z}[\sqrt{2},\sqrt{3}]$ is also true, where ${\Bbb Z}[\alpha]$ denotes the smallest subring of ${\Bbb C}$ which contains $\alpha\in{\Bbb C}$. It suffices to know whether $\sqrt{2}\in{\Bbb Z}[\sqrt{2}+\sqrt{3}]$ is true or not. With some manipulation one can get $$ 2\sqrt{2}\in {\Bbb Z}[\sqrt{2}+\sqrt{3}].  $$ It seems no hope to get $f(x)\in\Bbb{Z}[x]$ such that $$ f(\sqrt{2}+\sqrt{3})=\sqrt{2}, $$ which is equivalent to $\sqrt{2}\in{\Bbb Z}[\sqrt{2}+\sqrt{3}]$. But I don't have a proof. How should I go on?
100958
Show that $\sqrt{4 + 2\sqrt{3}} - \sqrt{3}$ is rational.   I've tried to attempt algebra on this problem. I noticed that there is some kind of nesting effect when trying to solve this. Please help me to understand how to attempt to denest this number.  Any help would be greatly appreciated.
94937
From a previous question, we have that $\sqrt[3]{3} \notin \mathbb{Q}(\sqrt[4]{3})$, and I am assuming that we would need to use this to justify the answer to the question.   Would it be right to use some idea of a splitting field here? Or is it something to do with $\mathbb{Q}(\sqrt[3]{3}, \sqrt[4]{3})$ not containing a primitive 3rd root of unity?   This is a past exam paper question worth 8 marks, so it doesn't seem like it's a few lined answer, but I am a little stuck on where to go.
41544
Denest $\sqrt {\sqrt[3]{5}-\sqrt[3]{4}}$.   I have tried completing square by several method but all failed. Can anyone help me please? Thank you.  p.s. I'm a poor question-tagger.
108948
I am trying to solve question 3.7 (b) from Chapter 15 in Artin's book "Algebra".  The problem is:     Is it true that $\sqrt[3]{5}\in \mathbb Q(\sqrt[3]{2})$?   It is clear by Eisenstein's criterion that both $\sqrt[3]{2}$ and $\sqrt[3]{5}$ have their minimal polynomails of egree $3$ over $\mathbb Q$.  Thus if we assume that $\sqrt[3]{5}\in \mathbb Q(\sqrt[3]{2})$, then we must have $\mathbb Q(\sqrt[3]{5})=\mathbb Q(\sqrt[3]{2})$.  This didn't get me anywhere.  Then I tried the most simple minded approach.  Say $\sqrt[3]{5}=a+b\sqrt[3]{2}+c\sqrt[3]{4}$, for some $a,b,c\in \mathbb Q$.  By cubing and rearranging, it is likely that we will arrive at some absurdity.  But clearly this shouldn't be what the author must have intended.  Can somebody point me towards a slick solution?
42887
$$ \left\lfloor\frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} +\dots+ \frac{1}{\sqrt{100}}\right\rfloor =\,? $$   I rationalized the denominator and then I think I should somehow group the numbers, but i don't know how.  Thanks in advance!
87058
This is from Spivak.  Prove that $\sqrt{2}+\sqrt{3}$ is irrational.  So far, I have this:  If $\sqrt{2}+\sqrt{3}$ is rational, then it can be written as $\frac{p}{q}$ with integral $p, q$ and in lowest terms.   $$\sqrt{2}+\sqrt{3}=\frac{p}{q}$$ $$2\sqrt{6}+5 =\frac{p^2}{q^2}$$ $$(2\sqrt{6}+5)q^2=p^2$$  And that's about where I get stuck. In a similar question (prove that $\sqrt{2}+\sqrt{6}$ is irrational,) I was able to show that both $p,q$ had to be even which is impossible. I obviously can't apply this trick here. Any hints?
119107
The answer to  hinged on the fact that $\sqrt[3]{35 + 18i\sqrt{3}} + \sqrt[3]{35 - 18i\sqrt{3}} = 7$. How does one evaluate such expressions?
111823
How can I find the roots of $x^3-3x+1$ using Cardano's formula?  So far, I found that   $$x = \sqrt[3]{\dfrac{-1+\sqrt{-3}}{2}} + \sqrt[3]{\dfrac{-1-\sqrt{-3}}{2}}$$ $$x = \sqrt[3]{\dfrac{-1+i\sqrt{3}}{2}} + \sqrt[3]{\dfrac{-1-i\sqrt{3}}{2}}$$ $$x = \sqrt[3]{\frac{-1}{2}+\frac{i\sqrt{-3}}{2}} + \sqrt[3]{\frac{-1}{2}-\frac{i\sqrt{-3}}{2}}$$  I am now trying to express each cubic radicand in their exponential form.  Euler's formula : $e^{i\theta} = \cos \theta + i\sin \theta$.  I have $r = |z| = 1$.  However, when I try to find the angle, I get  $$\theta = \arccos \frac{-1}{2} = 2\pi/3$$  but at the same time  $$\theta = \arcsin \frac{\sqrt{-3}}{2} = \pi/3$$  Shouldn't the two angles be the same for one radicand?  Plus, once I will have expressed the two radicands in their exponential form, what is the next step?  $$x = \sqrt[3]{e^{i2\pi/3}} + \sqrt[3]{e^{i4\pi/3}}$$ $$x = e^{i2\pi/9} + e^{i4\pi/9}$$  Am I on the right track?
80838
so I stumbled upon this equation/formula, and I have no idea how to prove it. I don't know how should I approach it: $$ \sqrt{2 + \sqrt{2 + \cdots + \sqrt{2 + \sqrt{\vphantom{\large A}2\,}\,}\,}\,}\ =\ 2\cos\left(\vphantom{\Large A}\pi \over 2^{n + 1}\right) $$  where $n\in\mathbb N$ and the square root sign appears $n$-times.  I thought about using sequences and limits, to express the LHS as a recurrence relation but I didn't get anywhere.  edit: Solved, thanks for your answers and comments.
52869
Questions:    $\cos\dfrac{\pi}{7}\cos\dfrac{2\pi}{7}\cos\dfrac{3\pi}{7}=?$ $\sin\dfrac{\pi}{7}\sin\dfrac{2\pi}{7}\sin\dfrac{3\pi}{7}=?$ $\tan\dfrac{\pi}{7}\tan\dfrac{2\pi}{7}\tan\dfrac{3\pi}{7}=?$   Approach:  If $x=\dfrac{\pi}{7}\implies 7x=\pi\implies 4x=\pi-3x\implies \sin 4x = \sin 3x$.Upon expansion, $$8cos^3x-4cos²x-4cosx+1=0$$     "$\cos \dfrac{\pi}{7}, \cos \dfrac{3\pi}{7}, \cos \dfrac{5\pi}{7}$will satisfy this cubic."   Note: I'm not able to understand the above line. Why do these values satisfy this cubic and how did we get these values?   These are therefore the roots of the cubic, and their product is:  $$\cos\dfrac{\pi}{7}\cos\dfrac{3\pi}{7}\cos\dfrac{5\pi}{7}=\dfrac{-1}{8}$$  Finally, $\cos\dfrac{5\pi}{7} = \cos\left(\pi-\dfrac{2\pi}{7}\right)=-\cos \dfrac{2\pi}{7}$. Replacing this in (1) gives $$\cos\dfrac{\pi}{7}\cos\dfrac{2\pi}{7}\cos\dfrac{3\pi}{7}=\dfrac{1}{8}$$  How can we apply this concept to find the values of    $\sin\dfrac{\pi}{7}\sin\dfrac{2\pi}{7}\sin\dfrac{3\pi}{7}?$ $\tan\dfrac{\pi}{7}\tan\dfrac{2\pi}{7}\tan\dfrac{3\pi}{7}?$   I'm not able to form the equations for these two problems. Please Help. .
47577
Prove by contradiction (not using a calculator) that $\sqrt6 + \sqrt2 &lt; \sqrt{15}$.   How do you approach such a problem?  I need to admit that I'm completely new to proof writing and I have completely no experience in answering that kind of questions. I tried to square it as I read online but I'm stuck with having $\sqrt3$ and not knowing how to get rid of it to have the answer as clear as the sun. squaring them yielded: $(\sqrt2 + \sqrt6)^2 =  8 + 4\sqrt3 &lt; 15$ which indeed holds but the problem is that I am not allowed to use calculator to check the little difference.  My final answer was $4\sqrt3 &lt; 7$.
42652
Since the formula is $a_n = a_1 + (n-1)d$ I could make $a_1 = 34_a, a_2 = 63_a$ and $a_3 = 112_a$. So what I need is a base a where the difference between $112_a$ and $63_a$ is equal to the difference between 63_a and 34_a. So I did:  $$1*a^2+1*a+2 - 6*a + 3 = 6*a+3 - 3a +4 \iff\\ a^2+a+2-a6+3 = a6+3-a3+4 \iff \\ a^2-5a+5=3a+7\iff \\ a^2-7a-2=0$$  This doesn't give me an integer solution. What went wrong?