Ta có:

`x^3-7x-12=3x root(3)(9(3+x-2x^2))`

Đặt:

`A=x`

`B=root(3)(2x-3)`

`C= -root(3)(9(x+1))`

Ta có:

`A^3+B^3+C^3=x^3+(2x-3)+[-9(x+1)]=x^3-7x-12`

`3ABC=3x root(3)(2x-3)[-root(3)(9(x+1))]= -3x root(3)(-9(2x-3)(x+1))=3x root(3)(9(3+x-2x^2))`

Phương trình trở thành:

`A^3+B^3+C^3=3ABC`

`A^3+B^3+C^3-3ABC=0`

`1/2(A+B+C)[(A-B)^2+(B-C)^2+(C-A)^2]=0`

`->[ (A+B+C=0), (A=B=C) :}`

Th1: `A=B=C`

`->{ (x=root(3)(2x-3)), (x= -root(3)(9(x+1))) :} ->{ (x^3=2x-3), (x^3= -9x-9) :}`

`->2x-3= -9x-9`

`->11x= -6`

`->x= -6/11`

Thay `x= -6/11` vào `x^3=2x-3`:

`(-6/11)^3=2(-6/11)-3` (vô lí)

Th2: `A+B+C=0`

`->x+root(3)(2x-3)-root(3)(9(x+1))=0`

`(x-2)+(root(3)(2x-3)-1)-(root(3)(9(x+1))-3)=0`

`(x-2)+[2(x-2)]/[root(3)((2x-3)^2)+root(3)(2x-3)+1]-[9(x-2)]/[root(3)(81(x+1)^2)+3root(3)(9(x+1))+9]=0`

`(x-2)(1+(2)/[root(3)((2x-3)^2)+root(3)(2x-3)+1]-(9)/[root(3)(81(x+1)^2)+3root(3)(9(x+1))+9])=0`

Vì `1+(2)/[root(3)((2x-3)^2)+root(3)(2x-3)+1]-(9)/[root(3)(81(x+1)^2)+3root(3)(9(x+1))+9]>0`

`->x-2=0`

`->x=2`

Vậy nghiệm của phương trình là `x=2`