`a)`

`(x+3)^3-x(3x+1)^2+(2x+1)(4x^2-2x+1)=28`

`x^3+9x^2+27x+27-x(9x^2+6x+1)+8x^3+1=28`

`x^3+9x^2+27x+27-9x^3-6x^2-x+8x^3+1=28`

`3x^2+26x+28=28`

`3x^2+26x=0`

`x(3x+26)=0`

`x=0` hoặc `3x+26=0`

`x=0` hoặc `x=-26/3`

Vậy `x in {0;-26/3}`

`b)`

`(x^2-1)^3-(x^4+x^2+1)(x^2-1)=0`

`x^6-3x^4+3x^2-1-(x^6-1)=0`

`x^6-3x^4+3x^2-1-x^6+1=0`

`-3x^4+3x^2=0`

`-3(x^4-x^2)=0`

`x^4-x^2=0`

`x^2(x^2-1)=0`

`x^2=0` hoặc `x^2-1=0`

`x=0` hoặc `x=1` hoặc `x=-1`

Vậy `x in {0;1;-1}`