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Menon"3,4,5 àŽŽàŽš àŽš àŽµ àŽµàŽ¶àŽ™ àŽ™àŽ³ àŽ†àŽ¯ àŽµàŽ° àŽš àŽš àŽ®àŽŸ àŽŸ àŽ€ àŽ° àŽ• àŽ£àŽ€ àŽ€ àŽš àŽ± Semi Perimeter 6 àŽ†àŽ£àŽ² àŽ² (3^3+4^3+5^3 = 6^3)6,8,10 àŽŽàŽš àŽš àŽµ àŽµàŽ¶àŽ™ àŽ™àŽ³ àŽ†àŽ¯ àŽµàŽ° àŽš àŽš àŽ®àŽŸ àŽŸ àŽ€ àŽ° àŽ• àŽ£àŽ€ àŽ€ àŽš àŽ± Semi Perimeter 12 àŽ†àŽ£àŽ² àŽ² (6^3+8^3+10^3 = 12^39,12,15 àŽŽàŽš àŽš àŽµ àŽµàŽ¶àŽ™ àŽ™àŽ³ àŽ†àŽ¯ àŽµàŽ° àŽš àŽš àŽ®àŽŸ àŽŸ àŽ€ àŽ° àŽ• àŽ£àŽ€ àŽ€ àŽš àŽ± Semi Perimeter 18 àŽ†àŽ£àŽ² àŽ² (9^3+12^3+15^3 = 18^3)àŽˆ àŽ• àŽ£ àŽš àŽš àŽ¬àŽš àŽ§àŽ€ àŽ€ àŽš àŽŽàŽš àŽ€ àŽ™ àŽ• àŽ² àŽ’àŽ° àŽ• àŽ°àŽ£ àŽšàŽ² àŽ• àŽ• àŽµ àŽš àŽªàŽ± àŽ± àŽ® àŽŽàŽš àŽš àŽ° àŽž àŽ¶àŽ¯ àŽ‰àŽ£ àŽŸ "May I approach the sottauiin 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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