If
(√6 + √2)9 A + B√C
---------- = -------
(√8)9 D
Then, find the integers values of A, B, C, and D with C square-free.
This is essentially a harder than average high school algebra TEXTBOOK PROBLEM! No computers needed!
(sqrt[6]+sqrt[2])^9 / (sqrt[8])^9 = ((sqrt[6]+sqrt[2])/sqrt[8])^9
sqrt[6]+sqrt[2] = sqrt[2]*(sqrt[3]+1) and sqrt[8] = 2*sqrt[2]
Then ((sqrt[6]+sqrt[2])/sqrt[8])^9 = ((sqrt[3]+1)/2)^9
Now break down the exponent
((sqrt[3]+1)/2)^9 = (sqrt[3]+1)^9/2^9 = ((sqrt[3]+1)^3)^3/2^9
(sqrt[3]+1)^3 = (sqrt[3]+1)^2*(sqrt[3]+1)
= (4+2*sqrt[3])*(sqrt[3]+1) = 2*(2+sqrt[3])*(1+sqrt[3])
= 2*(5+3*sqrt[3])
Then ((sqrt[3]+1)^3)^3/2^9 = (2*(5+3*sqrt[3]))^3/2^9
= (5+3*sqrt[3])^3*2^3/2^9 = (5+3*sqrt[3])^3/2^6
(5+3*sqrt[3])^3 = (5+3*sqrt[3])^2*(5+3*sqrt[3]) = (52+30*sqrt[3])*(5+3*sqrt[3]) = 530+306*sqrt[3]
Then (5+3*sqrt[3])^3/2^6 = (530+306*sqrt[3])/2^6
= 2*(265+153*sqrt[3])/2^6 = (265+153*sqrt[3])/2^5
= (265+153*sqrt[3])/32.
(sqrt[6]+sqrt[2])^9 / (sqrt[8])^9 = (265+153*sqrt[3])/32.
A=265, B=153, C=3, D=32.
Solution
