高中数学，化简这两个试子。（不懂的别乱答）

换元法

第一个，
上下同时乘以a^(1/3)
你会发现，分母变成了一个立方差的展开式
[4b^(2/3)+2(ab)^(1/3)+a^(2/3)][a^(1/3)-2b^(1/3)]=a-8b
分子是a^(1/3)(a-8b)a^(2/3)=a(a-8b)

所以原式=a

第二个
x-1=[x^(1/3)-1] [x^(2/3)+x^(1/3)+1]
所以第一项，(x-1)/[x^(2/3)+x^(1/3)+1]=x^(1/3)-1

x+1=[x^(1/3)+1] [x^(2/3)-x^(1/3)+1]
所以第二项，(x+1)/[x^(1/3)+1]]=x^(2/3)-x^(1/3)+1

所以
原式=x^(1/3)-1+x^(2/3)-x^(1/3)+1-[x-x^(1/3)]/[x^(1/3)-1]
=x^(2/3)-[x-x^(1/3)]/[x^(1/3)-1]
=[x-x^(2/3)-x+x^(1/3)]/[x^(1/3)-1]
=[-x^(2/3)+x^(1/3)]/[x^(1/3)-1]
= -x^(1/3)

第一个式子
设a^(1/3)=A，b^(1/3)=B，则：a=A^3，b=B^3，代入原式中：
原式
=(A^4-8AB^3)*A/[(4B^2+2AB+A^2)(1-2B/A)]
=A^2(A^3-8B^3)/[(4B^2+2AB+A^2)(A-2B)/A]
=A^3[A^3-(2B)^3]/[(A-2B)(A^2+2AB+4B^2)]

=A^3[A^3-(2B)^3]/[A^3-(2B)^3]
=A^3
=a

第二个式子
设x^(1/3)=X，则x=X^3
原式
=(X^3-1)/(X^2+X+1)+(X^3+1)/(X+1)-(X^3-X)/(X-1)
=(X-1)(X^2+X+1)/(X^2+X+1)+(X+1)(X^2-X+1)/(X+1)-X(X^2-1)/(X-1)
=(X-1)+(X^2-X+1)-X(X+1)(X-1)/(X-1)
=X-1+X^2-X+1-X(X+1)
=X^2-X^2-X
=-X
=-x^(1/3)

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