若x+1/x=a，求x六次方+1/x六次方

x^6+(1/x^6)
=[x^2+(1/x^2)]*[x^4+(1/x^4)-1]
由x+(1/x)=a
则[x+(1/x)]^2=a^2
x^2+(1/x^2)+2=a^2 x^2+(1/x^2)=a^2-2
[x^2+(1/x^2)]^2=(a^2-2)^2
x^4+(1/x^4)+2=a^4-4a^4+4
x^4+(1/x^4)=a^4-4a^2+2
则原式=[x^2+(1/x^2)]*[x^4+(1/x^4)-1]
=[a^2-2]*[a^4-4a^2+1]
=a^6-6a^4+9a^2-2

(xxxxx+ 1/xxxxxx)
=
(xx + 1/xx ) ( xxxx + 1/xxxx - 1)
=(aa - 2 )((xx + 1/xx)^2 - 3)
=(aa-2)((aa-2)^2 - 3)
=(aa-2)(aaaa - 4aa + 1)
=a^6 - 6a^4 + 9aa -2

a6+2a4-7a2-2

用上完全平方公式和立方和公式哟

先对条件两边平方，再对所求式子进行立方和公式。
对x四次方+1/x四次方，用上完全平方公式，OK？