# Cube root and short multiplication formulas [closed]

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! :)

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

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• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – B. Goddard, Xam, Lord Shark the Unknown, Claude Leibovici, Paramanand Singh
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• What is $(1+\sqrt2)^3$? – Lord Shark the Unknown Sep 13 '17 at 19:30
• @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). – Xander Henderson Sep 13 '17 at 19:38
• 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. – Aras Sep 13 '17 at 20:14

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 $x^3+3x-14=0\,$.

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

Hint:

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

• I can solve it from here, but could you explain how you get to this point exactly? Thanks a lot! – Aras Sep 13 '17 at 20:04
• 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. – Bernard Sep 13 '17 at 20:09
• Oh so basically still using a^2+2ab+b^2, just the cube variant of it? – Aras Sep 13 '17 at 20:13
• There's a mid-school formula for $(a+b)^3$: it's equal to $a^3+3a^2b+3ab^2+b^3$. – Bernard Sep 13 '17 at 20:14
• Yup, got that, thanks a lot! :) – Aras Sep 13 '17 at 20:15

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.

• @gt6989b Yes, it's the only real solution, thanks, I edited :) – john doe Sep 13 '17 at 19:43
• Could you explain a little more on how you get -3x = x^3, before solving for (a-b)^3? Thanks! – Aras Sep 13 '17 at 20:31
• @Aras see my edited answer. Hopefully, you will understand it clearly – john doe Sep 13 '17 at 21:18
• 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! – Aras Sep 14 '17 at 12:42

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

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

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

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