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 June 5th, 2017, 12:14 AM #1 Member   Joined: Aug 2014 From: Lithuania Posts: 60 Thanks: 3 simple algebra question Find the value of $\displaystyle x^3-6x$ when $\displaystyle x=(20+14\sqrt{2})^{1/3}+(20-14\sqrt{2})^{1/3}$. Here's my attempt, please check whether it's correct. Last edited by skipjack; June 5th, 2017 at 12:56 AM.
 June 5th, 2017, 12:42 AM #2 Senior Member   Joined: Feb 2016 From: Australia Posts: 1,597 Thanks: 546 Math Focus: Yet to find out.
 June 5th, 2017, 12:54 AM #3 Global Moderator   Joined: Dec 2006 Posts: 19,062 Thanks: 1619 As $\displaystyle x=(20+14\sqrt{2})^{1/3} + (20-14\sqrt{2})^{1/3} = 2 + \sqrt2 + 2 - \sqrt2 = 4$, the answer = 4³ - 24 = 40.
 June 5th, 2017, 11:50 PM #4 Member   Joined: Aug 2014 From: Lithuania Posts: 60 Thanks: 3 How did you get that $\displaystyle \sqrt[3]{20+14\sqrt{2}}=2+\sqrt{2}$
June 6th, 2017, 03:27 AM   #5
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Quote:
 Originally Posted by kaspis245 How did you get that $\displaystyle \sqrt[3]{20+14\sqrt{2}}=2+\sqrt{2}$
I'm not certain how SkipJack worked it out, but one way is to assert that

$\displaystyle (20+14\sqrt{2})^{1/3} = u + v\sqrt{2}$

where u and v are positive constants. Then...

$\displaystyle 20 + 14\sqrt{2} = (u+v\sqrt{2})^3 = u^3 + 3\sqrt{2}u^2v + 6 u v^2 + 2\sqrt{2} v^3$

Therefore, using coefficient matching,

$\displaystyle u^3 + 6 uv^2 = 20$ and
$\displaystyle 3u^2v + 2v^3 = 14$.

Then these simultaneous equations can be solved using your method of choice. One can note that when substituting u = 1, u = 2, u = 3... etc. into the first equation to attempt to find the interval with which the solution lies, a rather fortunate circumstance appears in that a pair of solutions u = 2, v = 1 reveals itself, so once it's confirmed that u = 2, v=1 works also for the second equation, you can then simplify the original problem.

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