Is there a really quick way of showing that:
$$\sqrt[3]{49-25\sqrt{2}}$$
Can be written in the form:
$$a+b\sqrt{2}$$
Is there a way to generalize which integers $a$ and $b$ can be rewritten in such a way?
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3Why do you think it is possible to write that cubic root in that way? – DonAntonio May 07 '14 at 03:45
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@DonAntonio I was able to do the algebra out (ie setting the expression equal to a+bsqrt2 and setting parts equal to one another.) I was also given the hint that a and b are integers so I was able to do some factoring along the way. – Jackson May 07 '14 at 03:49
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One way is to notice that the minimum polynomial has degree $2$, so there's at least hope for this to be true; if the degree was higher than $2$, it's not possible. – May 07 '14 at 03:51
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Then what are you asking, @Jackson: how to do it "quickly"? Perhaps you show your work to know what'd be "quickly" in this case... – DonAntonio May 07 '14 at 03:52
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@Don Because it is, viz. $\ 3 -\sqrt{2}\ \ $ – Bill Dubuque May 07 '14 at 03:52
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That I know, @BillDubuque: the question is how the OP suspected, or knew, that it is so. – DonAntonio May 07 '14 at 03:53
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1@DonAntonio I was wondering if there was some sort of characteristic of these numbers that guaranteed this outcome. I solved it the standard algebra way (45=a^3-6ab^2 and 29=3a^2b+2b^3). Since I was given in the problem that a and b were integers, I used the fact that 29 is prime to determine a and b. – Jackson May 07 '14 at 04:00
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@Jackson See my answer below for a generalization. – May 07 '14 at 04:08
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possible duplicate of How does one evaluate $\sqrt{x + y} + \sqrt{x - y}$? – mercio May 07 '14 at 08:41
1 Answers
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If we assume $\sqrt[3]{45 - 29 \sqrt{2}} = a + b \sqrt{2}$ where $a$ and $b$ are integers, then cubing both sides and expanding gives us $a^3 + 3 a^2 b \sqrt{2} + 6 a b^2 + 2 b^3 \sqrt{2}$, so we want to solve $$\eqalign{a^3 + 6 a b^2 &= 45\cr 3 a^2 b + 2 b^3 &= -29\cr}$$ Now $b$ divides $3 a^2 b + 2 b^3$ and $29$ is prime, so $b$ must be $\pm 1$. Moreover $b = +1$ would make $3 a^2 + 2 b^3 > 0$, so we must have $b = -1$. Then the second equation becomes $-3 a^2 - 2 = -29$ or $a^2 = 9$, so $a = \pm 3$. Finally the first equation becomes $\pm 27 \pm 18 = 45$ which is true for $a=+3$. Thus the answer is $3 - \sqrt{2}$.
EDIT: The problem keeps getting edited, but here are some more general considerations. For $\sqrt[3]{A + B \sqrt{m}} = a + b \sqrt{m}$ where $m$ is a square-free integer, you need $$ \eqalign{ a (a^2 + 3 m b^2) &= A\cr b (3 a^2 + m b^2) &= B\cr}$$ so $a$ divides $A$ and $b$ divides $B$. That reduces you to finitely many possibilities. To cut it down further you might note that $a \equiv A \mod 3$ and $mb\equiv B \mod 3$, and (if $m > 0$) $a$ and $b$ have the same signs as $A$ and $B$ respectively. Also, $$ A^2 - B^2 m = (a^2 - b^2 m)^3$$ so a necessary condition for this to work is that $A^2 - B^2 m$ is the cube of an integer. For example, $49^2 - 2 \times 25^2 = 1151 $ is not the cube of an integer, so $49 - 25 \sqrt{2}$ won't work.
Robert Israel
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The OP already solved it that way, see the comments to the question. He's asking about generalizations. – Bill Dubuque May 07 '14 at 04:05
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I see the OP has now changed the question to $45 + 25 \sqrt{2}$. This one does not simplify. – Robert Israel May 07 '14 at 04:06
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@Robert Sorry, I think somebody else mistakenly edited it. Good catch. – Jackson May 07 '14 at 04:07
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@RobertIsrael Yes I fixed it back. At some point, somebody even edited to say a+sqrtb, which is clearly wrong. – Jackson May 07 '14 at 04:10
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How do you know that $b\equiv B\pmod 3$? Could you show me the proof? Thanks. – user26486 May 07 '14 at 11:09
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@mathh $\ {\rm mod}\ 3!:\ A = a^3 + 3mab^2 \equiv a^3 \equiv a\ $ by little Fermat. Similarly $\ B\equiv mb\ $ (originally it mistakenly read $\ B\equiv b,\ $ so maybe that caused confusion). – Bill Dubuque May 07 '14 at 15:27