3 ^ (11) + 7 ^ (11) = c ^ (2) - b ^ (2)
give 11 sets at least for c and b.
rational numbers of course.
2006-07-20 03:49:42 · 6 answers · asked by rajesh bhowmick 2 in Science & Mathematics ➔ Mathematics
ok ffd now give the formula for natural numbers.
2006-07-20 04:23:05 · update #1
3^11+7^11
= 1977503890
c^2-b^2 = 1977503890
(c-b)(c+b) = 1977503890
For solutions, we can let:
c - b = 1, c + b = 1977503890
c - b = 2, c + b = 1977503890/2
Or in general if we let c-b = k, c+b = 1977503890/k, then
we have solutions:
c = 988751945/k + k/2
b = 988751945/k - k/2
And this is rational for any rational k.
2006-07-20 04:16:32 · answer #1 · answered by fatal_flaw_death 3 · 1⤊ 0⤋
Substitute values for b & find the corresponding value for c. Try to get a scientific calculator
2006-07-20 10:58:28 · answer #2 · answered by ashok 1 · 0⤊ 0⤋
C=3955362074 and b=1977681037
2006-07-20 11:53:01 · answer #3 · answered by Sharik 2 · 0⤊ 0⤋
sorry buddy, not doing your math homework for you, hire a tutor
2006-07-20 10:53:24 · answer #4 · answered by NBGirl 5 · 0⤊ 0⤋
Do your own homework!
2006-07-20 10:52:44 · answer #5 · answered by Ashlee S 4 · 0⤊ 0⤋
-INFINITY
I am sure the answer lies in this interval if not ur qs is wrong
2006-07-20 10:57:18 · answer #6 · answered by Anonymous · 0⤊ 0⤋