Gig Info:Date: 15.06.2009 Location: Secession, Vienna, AustriaAnother fascinating night at the secession! This time brother rain did not visit us, but some other old friends or foes, beginning with pol and ending with ice, did. In the name of art they left without harmin anyone. What a great show!Comment by Eric, Date: 25.11.2015 08:48:00 Finndig this post solves a problem for me. Thanks! Comment by Bassam, Date: 26.11.2015 15:34:00 I was so confused about what to buy, but this makes it <a href="http://exrujbwxxnm.com">unednstardable.</a> Comment by Nobi, Date: 26.11.2015 21:51:00 Holy cocinse data batman. Lol! http://amhrnfmfcf.com [url=http://zbiqjn.com]zbiqjn[/url] [link=http://pjozqj.com]pjozqj[/link] Comment by Rizen, Date: 28.11.2015 09:02:00 ர ம ச ம // கட ர ண ட ந ள ல வ ? சர அப ப ம ன வத ந ள கட த றக கல ன ன , ஏலத த ல வ டப பட ம ... //ஆஹ ... ச ல ல மறந த ட ட ன ... த ட ர ன ம ட சர ய ல லன ன ல வ extend ஆக ம ...// க தல , கத த ர க க ன ன எத ய வத பண ண ங க... ஆன கவ த க வ தன ன ட ர ச சர பண ண அப ப றம இர க க .........! //உங க எல ல ர ட நல ல ந ரம இத வர க க ம அப பட எத வ ம வந த த ல க கல ... Comment by Hasan, Date: 01.12.2015 03:00:00 Menon"3,4,5 എന ന വ വശങ ങള ആയ വര ന ന മട ട ത ര ക ണത ത ന റ Semi Perimeter 6 ആണല ല (3^3+4^3+5^3 = 6^3)6,8,10 എന ന വ വശങ ങള ആയ വര ന ന മട ട ത ര ക ണത ത ന റ Semi Perimeter 12 ആണല ല (6^3+8^3+10^3 = 12^39,12,15 എന ന വ വശങ ങള ആയ വര ന ന മട ട ത ര ക ണത ത ന റ Semi Perimeter 18 ആണല ല (9^3+12^3+15^3 = 18^3)ഈ ക ണ ന ന ബന ധത ത ന എന ത ങ ക ല ഒര ക രണ നല ക ക വ ന പറ റ മ എന ന ര സ ശയ ഉണ ട "May I approach the saiuitton as follows:3, 4 and 5 is a Pythagorean triplet.3^3 +4^3 + 5^3 = 27 + 64 + 125=216Also [(3+4+5)/2]^3 = 6^3 =216It proves that 3^3 +4^3 + 5^3 = [(3+4+5)/2]^3We notice that each successive Pythagorean triplets that you have mentioned in your post is an integral multiple of the triplet (3,4,5) and hence a general Pythagorean triplet in your post can be expressed as (3k, 4k, 5k)where k= 1, 2, 3, 4,.........Now (3k)^3 + (4k)^3 + (5k)^3 =k^3(3^3 +4^3+ 5^3)= k^3 [(3+4+5)/2]^3 by substitution from the previous relationHence, (3k)^3 + (4k)^3 + (5k)^3 =k^3 [(3+4+5)/2]^3 = [k(3+4+5)/2]^3 = [(3k+4k+5k)/2]^3i.e (3k)^3 + (4k)^3 + (5k)^3 = [(3k+4k+5k)/2]^3In the above relation put k=1, 2, 3, 4, 5..... Then we get all those relations mentioned in your post.This is a special property of Pythagorean triplets of the form (3k,4k,5k)If you examine the Pythagorean triplet (5,12,13)we see that this property namely sum of cubes equals cube of semi-perimeter does not hold. http://mztbeecgyi.com [url=http://jluuyt.com]jluuyt[/url] [link=http://jpzfklx.com]jpzfklx[/link] Comment by Mark, Date: 12.05.2016 12:35:00 IhyGF9 http://www.y7YwKx7Pm6OnyJvolbcwrWdoEnRF29pb.com Post Comment |