`a)`
`(x-2)^3-(x-3)(x^2+3x+9)+6(x+1)^2=15`
`\to x^3-6x^2+12x-8-(x^3-27)+6(x^2+2x+1)=15`
`\to x^3-6x^2+12x-8-x^3+27+6x^2+12x+6=15`
`\to 24x+25=15`
`\to 24x=-10`
`\to x=-5/12`
Vậy `x=-5/12`
``
`b)`
`6x^2-6x(-2+x)=36`
`\to 6x^2+12x-6x^2=36`
`\to 12x=36`
`\to x=3`
Vậy `x=3`