Wolfram alpha fails to answer this (If you type it directly)

#brainexercise #matholympics
Wolfram alpha fails to answer this (If you type it directly)

Comment your answer in the comment section down below

[sarantiskalaitzis]

If we don't convert the original in the form a+b=3 and let it be as a-b=3 then we must use the product and calculate a^3-b^3 from also (a^2+b^2)= (a-b)^2+2ab and then a^4+b^4= (a^2+b^2)^2- 2•a^2•b^2 etc ....but calculations having a+b instead of a-b, are easier.

[sarantiskalaitzis]

🎉I think it's easier to use that (a^3+b^3)•(a^4+b^4)= a^7+a^3•b^4+b^3•a^4+b^7 =a^7+b^7+a^3•b^3•(a+b)= a^7+b^7+(ab)^3•(a+b)= 129+3•(ab)^3 "(1)". Then we recall that a+b=3 so 3ab•3 =27-9ab. "(2)". Also a^4+b^4= (a^2+b^2)^2-2•a^2•b^2= [(a+b)^2-2•ab]^2-2(ab)^2= 129+ (9-2ab)^2-2(ab)^2. Let ab=t. Then (1) becomes (27-9t)•[(9-2t)^2-2t^2)= 129+3•t^3. Finally we get 3rd grade equation ..
t^3- 18t^2+81t-98 =0. Etc..

[fhffhff]

{c=4-a-b, a=1-b/2±√(-3/4b²+b-2),
0=(b³-3b²+2, 5b-15, 5)²+25, 75(b-125/103)²-11760/103{127, 45±26, 51i}

[raghvendrasingh]

❤ nice problem
let a^7 = x+64 & b^7 = 65 - x
a+b = 3, a^7+b^7 = 129
let us consider the identity
(a+b)^7 - a^7 - b^7 = 7ab(a+b)(a°2+ab+b^2)^2
3^7 - 129 = 21t (9 - t)^2 { where t = ab also (a+b)^2 = 3^2 gives a^2+b^2 = 9 - 2t }
3^6 - 43 = 7t (t^2 - 18 t+ 81)
t^3 - 18t^2+81t - 98 = 0
(t - 2)(t^2 - 16t+49) = 0
t = 2, 8+/- √15
a+b = 3, ab = 2 gives a = 2, b = 1 or a = 1, b= 2
a^7 = x+64 hence
x = 64, - 63
from second case of ab we get imaginary roots.

[VijayKumar-rwpt]

(2022)^ 22 divided by 7 find remainder

Please solve this