✔ 最佳答案
(b) Multiply all terms by [ sqrt 3 + sqrt 2]^6, we get
[sqrt 3 + sqrt 2]^6[sqrt3 + sqrt 2]^6 + [sqrt 3 - sqrt 2]^6[sqrt 3 + sqrt 2]^6 = 970[sqrt 3 + sqrt 2]^6
[sqrt 3 + sqrt 2]^6[sqrt 3 + sqrt 2]^6 + 1 = 970[sqrt 3 + sqrt 2]^6
Let [sqrt 3 + sqrt 2]^6 be x, the equation becomes
x^2 + 1 = 970x
x^2 - 970x + 1 = 0
Solving this quadratic equation, we get
x = [ sqrt 3 + sqrt 2]^6 = 485 + 198 sqrt 6. ( 485 - 198 sqrt 6 is rejected since it equals to zero.)
Note : Putting n = 1 in part (a) is not exactly correct, you should use the relation (a^n)(b^n) = (ab)^n
2012-08-25 09:05:55 補充：
Correction : 485 - 198 sqrt 6 equals to - 0.00103, not exactly equals to 0. It is rejected because it is negative.
2012-08-25 09:10:14 補充：
Sorry. 485 - 198 sqrt 6 is not negative, it is 0.00103. So reject because too small.
2012-08-25 11:16:16 補充：
Note : 485 - 198 sqrt 6 is the answer for [sqrt 3 - sqrt 2]^6.
