20070618, 22:36  #1 
May 2004
New York City
2·2,099 Posts 
Solve this equation
Solve for x:
(x+9)^(1/3)  (x9)^(1/3) = 3 (This may be too easy for this forum, but I've always liked the solution method.) 
20070618, 23:23  #2 
∂^{2}ω=0
Sep 2002
República de California
10011000111111_{2} Posts 
x[sup]2[/sup]=80
Last fiddled with by ewmayer on 20070618 at 23:23 
20070619, 07:50  #3 
"Lucan"
Dec 2006
England
6,451 Posts 

20070619, 12:20  #4 
Bronze Medalist
Jan 2004
Mumbai,India
2^{2}×3^{3}×19 Posts 
Solution!

20070619, 12:56  #5 
Dec 2005
3·5·13 Posts 
Taking cubes on both sides gives the result without any noteworthy manipulation.
If there is an elegant solution or a smart substitution it still needs to be very short to beat the straightforward one 
20070619, 14:23  #6 
Jun 2007
Moscow,Russia
7·19 Posts 
Equation doesn't have solution(s)

20070619, 15:40  #7  
"Lucan"
Dec 2006
England
6,451 Posts 
Quote:
3[(x+9)^{1/3}(x9)^{1/3}](x^{2}81)^{1/3} = 9 at which point we use the original equation to substitute 3 for [...] David Last fiddled with by davieddy on 20070619 at 15:42 

20070619, 15:51  #8 
Account Deleted
"Tim Sorbera"
Aug 2006
San Antonio, TX USA
17·251 Posts 

20070619, 21:14  #9 
Aug 2005
Brazil
2·181 Posts 
Not quite.

20070620, 07:32  #10  
Bronze Medalist
Jan 2004
Mumbai,India
2^{2}×3^{3}×19 Posts 
Solutions!
Quote:
A good problem but not instructional enough. Its not good nor instructional getting straight solutions from the 'know all's'. One right answer as ewmayer's is enough to tackle the problem A hint from Keyes was good as it saved a lot of 'computation'. I have to use the term in these computer days as this is piffle for the comp. As usual Davieddy's 'solution' and as he termed it 'entertainment' made things more confused and you can all see for your selves. Imagine substituting a whole equation for an integer ? And he claims to be a math's tutor ! A little knowledge is a very dangerous thing. Arrogance is even worse! Now given that the solution is correct (I have not checked it out but accept Ewmayer's as correct) the next question is to generalise it from a particular problem. My question is how would you solve the problem when the exponents are not equal? Lets say they are p and q and not just 1/3 ? How would you get the rationalising factor esp. when there are three or more surds? Thank you Davar55 as it made me revise my algebra! BTW: if you have a better solution than keyes kindly respond and let me know. Mally Last fiddled with by mfgoode on 20070620 at 07:35 

20070620, 08:03  #11  
Aug 2004
2·5·13 Posts 
Quote:
Chris 

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