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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 $\displaystyle x^36x$ when $\displaystyle x=(20+14\sqrt{2})^{1/3}+(2014\sqrt{2})^{1/3}$. 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: 1,578 Thanks: 537 Math Focus: Yet to find out.  
June 5th, 2017, 12:54 AM  #3 
Global Moderator Joined: Dec 2006 Posts: 18,843 Thanks: 1565 
As $\displaystyle x=(20+14\sqrt{2})^{1/3} + (2014\sqrt{2})^{1/3} = 2 + \sqrt2 + 2  \sqrt2 = 4$, 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  
Senior Member Joined: Apr 2014 From: Glasgow Posts: 2,099 Thanks: 703 Math Focus: Physics, mathematical modelling, numerical and computational solutions  Quote:
$\displaystyle (20+14\sqrt{2})^{1/3} = u + v\sqrt{2}$ 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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