1) a^3 + b^3 = c^3 +d ^3

2) x^3 + y^3 = z^3

^ - power of

Find the a,b,c and d for (1), x,y and z for (2). Two separate questions

First equation
10^3 + 9^3 = 12^3 + 1^3 (=1729, known as Ramanujan number)
There can be as many solutions for the first equation...
(a) 167^3 + 436^3 = 228^3 + 423^3 = 255^3 + 414^3 (=87539319)
(b) 2421^3 + 19083^3 = 5436^3 + 18948^3 = 10200^3 + 18072^3 = 13322^3 + 16630^3 = (6963472309248)
(c) 38787^3 + 365757^3 = 107839^3 + 362753^3 = 205292^3 + 342952^3 = 221424^3 + 336588^3 = 231518^3 + 331954^3 (= 48988659276962496)
(d) 6^3 + (−5)^3 = 4^3 + 3^3 = (91)
(e) 6^3 + 8^3 = 9^3 + (-1)^3 = 12^3 + (-10)^3 (=728 )

[b]Second Equation...[/B]
Fermat's last theorem states that if n is a whole number strictly bigger than 2, then you can never find three whole numbers x, y and z such that x^n+y^n=z^n.
In other words, If an integer n is greater than 2, then a^n + b^n = c^n has no solutions in non-zero integers a, b, and c.
Hope, we don't have any solution for equation if Fermat was right...

HI Manoj , Me Agree with Ritesh

Answer 1. - A = 10 , B = 9 , C = 12, D = 1

10^3 + 9^3 = 12^3 + 1^3 (=1729, known as Ramanujan number)

Answer 2. - Not possible if u have answer plz give us.


Regards

Nikhil Rattan

Yeah iasritesh and nikhil you guys r great... you guyz got the answer... even i dont know about these many answers for question number (1)... i only know about Ramanujan number... where any formula for question 1 or its trial and error method...

For question 2 Fermat was correct... no solution for question (2)... again both of u r correct...

Thanks Manoj i was thinking

Same for Question 2 . That there should not be any solution for that.

As i saw the answer of ritesh i got sure about that.

Regards

Nikhil Rattan

Manojks,
As far my knowledge goes, there is no direct formula for equation 1.
Its a purely trial - error method. In higher level mathematics, we can have some facilities, so that the trial - error work could be reduced.