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I've got this math assignment. I'm stuck with it and I'd love it if someone could help me out with it, thanks! :)

$$\sqrt[3]{5\sqrt2 + 7} - \sqrt[3]{5\sqrt2-7}=2$$

Edit: So far I've tried using short multiplication formules, simplifying the roots or doing the opposite to suite the $a^2 + 2ab + b^2$, but so far I think I was aproaching this the wrong way. My bad for not following some rules, my first post here sorry. Edit 2: Got it all, explained it to the class. Huge thanks to everyone this community is really awesome! :)

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closed as off-topic by B. Goddard, Xam, Lord Shark the Unknown, Claude Leibovici, Paramanand Singh Sep 14 '17 at 7:33

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If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ What is $(1+\sqrt2)^3$? $\endgroup$ – Lord Shark the Unknown Sep 13 '17 at 19:30
  • $\begingroup$ @Aras Questions that take the form "Here is a problem I can't solve. Halp!" look like homework questions. Whether they are or are not is irrelevant. If you want to get more constructive help, I would suggest that you edit your post to include some context, such as anything that you have tried in order to solve the problem (whether they worked or not). $\endgroup$ – Xander Henderson Sep 13 '17 at 19:38
  • $\begingroup$ Yes, thanks for the critique, I fixed it a little bit, hopefully it's better :) Also thanks everyone for the help! This was really fast. $\endgroup$ – Aras Sep 13 '17 at 20:14
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Hint:  let $a=\sqrt[3]{5\sqrt2 + 7}, b =\sqrt[3]{5\sqrt2-7}$ then $a^3-b^3=14$ and $ab=1\,$, so:

$$(a-b)^3=a^3-b^3-3ab(a-b)=14-3(a-b)$$

Now look for the real solution(s) of the equation $x^3+3x-14=0\,$.

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    $\begingroup$ This is awesome, huge thanks! :) $\endgroup$ – Aras Sep 13 '17 at 20:03
  • $\begingroup$ I marked this as the answer because it seems to be the easiest / fastest choice. Thanks! $\endgroup$ – Aras Sep 14 '17 at 12:40
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Hint:

$$5\sqrt 2+7=(\sqrt2+1)^3.$$

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  • $\begingroup$ I can solve it from here, but could you explain how you get to this point exactly? Thanks a lot! $\endgroup$ – Aras Sep 13 '17 at 20:04
  • $\begingroup$ It is a classic trick from high school that some $a+b\sqrt n$ is actually $(x+y\sqrt n)^2$ for some $x, y$. It allows to simplify expressions like $\sqrt{11+6\sqrt2}$. So I wondered whether it were the expansion of some $(x+y\sqrt 2)^3$, since there was a cube root. $\endgroup$ – Bernard Sep 13 '17 at 20:09
  • $\begingroup$ Oh so basically still using a^2+2ab+b^2, just the cube variant of it? $\endgroup$ – Aras Sep 13 '17 at 20:13
  • $\begingroup$ There's a mid-school formula for $(a+b)^3$: it's equal to $a^3+3a^2b+3ab^2+b^3$. $\endgroup$ – Bernard Sep 13 '17 at 20:14
  • $\begingroup$ Yup, got that, thanks a lot! :) $\endgroup$ – Aras Sep 13 '17 at 20:15
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Here's how you can do it - take equation to be LHS $= x = a-b$. Then use identity $(a-b)^3 = a^3 - b^3 -3ab(a-b)$ where $a = (5\sqrt 2 +7)^{1/3}$ and $b= (5\sqrt 2 - 7)^{1/3} $. Then solving, $ a^3 - b^3 = (5\sqrt 2 +7) - (5\sqrt 2 - 7) = 14 , ab = (50 - 49)^{1/3} = 1^{1/3}=1 $ (apply $ (x+y)(x-y) = x^2 - y^2 )$ :- you will get $ x^{3}=14 - 3x $ whose only real solution is 2.

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  • $\begingroup$ @gt6989b Yes, it's the only real solution, thanks, I edited :) $\endgroup$ – john doe Sep 13 '17 at 19:43
  • $\begingroup$ Could you explain a little more on how you get -3x = x^3, before solving for (a-b)^3? Thanks! $\endgroup$ – Aras Sep 13 '17 at 20:31
  • $\begingroup$ @Aras see my edited answer. Hopefully, you will understand it clearly $\endgroup$ – john doe Sep 13 '17 at 21:18
  • $\begingroup$ Yes awesome thanks! But I marked the other answer as a solution, it was a bit simpler to understand, but your details were also very important. Thanks! $\endgroup$ – Aras Sep 14 '17 at 12:42
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Ask what $5\sqrt2 + 7$ and $5\sqrt2-7$ would look like as cubes of something. Envision that:

$$5\sqrt2 + 7=\left(a+b \sqrt{2}\right)^3$$

This implies that $$7+5\sqrt2=\left(a^3+6ab^2\right) + \left(3a^2b+2b^3\right)\sqrt{2}$$ So it must be that $$\begin{align} \begin{cases} 7&=a^3+6ab^2 &=a(a^2+6b^2) \\ 5&= 3a^2b+2b^3 &= b(3a^2+2b^2)\end{cases}\end{align}$$

a quick inspection, using the convenient fact that $7$ and $5$ are primes, leads us to $$[a=b=1] \implies $$ $$7+5\sqrt2 =\left(1+ \sqrt{2}\right)^3$$

Now try it for

$$5\sqrt2 - 7=\left(c+d \sqrt{2}\right)^3$$

You can bet that

$$[c=-1 \ \text{and} \ d=1] \implies$$

$$-7+5\sqrt2 =\left(-1+\sqrt{2}\right)^3$$

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  • $\begingroup$ Interesting approach, thanks! $\endgroup$ – Aras Sep 14 '17 at 12:44
  • $\begingroup$ Hey you're welcome $\endgroup$ – AmateurMathPirate Sep 14 '17 at 13:24

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