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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 situitaon 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.
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(VISITOR) AUTHOR'S NAME
FourtyGang
MESSAGE TIMESTAMP
17 december 2014, 03:12:44
AUTHOR'S IP LOGGED
62.210.78.179
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