Question

已知a+a分之1=3

已知a+a分之1=3,则a3次方+(a3次方)分之1等于几，a4次方+(a4次方)分1呢？

Answer 1

a+1/a=3
平方得a^2+1/a^2 = 7
a^3+1/a^3 = (a+1/a)(a^2-1+1/a^2) = 3 * ( 7 - 1) = 18
a^4+1/a^4 = (a^2+1/a^2)^2 - 2 = 7^2-2=47

Answer 2

a^3+1/a^3
=(a+1/a)(a^2+a*1/a+1/a^2)
=(a+1/a)[(a+1/a)^2-1]
=3x(3^2-1)
=24

a^4+1/a^4
=(a^2+1/a^2)^2-2
=[(a+1/a)^2-2]^2-2
=(3^2-2)^2-2
=49-2
=47

Answer 3

解：已知 a＋﹙1／a﹚＝3
∴[a＋﹙1／a﹚]²＝9
即 a²＋﹙1／a²﹚＋2＝9
∴a²＋﹙1／a²﹚＝7
∵﹙a^4＋a²＋1﹚／a²
原式＝﹙a^4＋a²＋a﹚÷a²
＝a²＋1＋﹙1／a²﹚
∴a²＋1＋﹙1／a²﹚＝7＋1＝8