`a)`
`(x-1)^3+(2-x)(4+2x+x^2)+3x(x+2)=16`
`\to (x^3-3x^2+3x-1)+(2^3-x^3)+(3x^2+6x)=16`
`\to x^3-3x^2+3x-1+8-x^3+3x^2+6x=16`
`\to (x^3-x^3)+(3x^2-3x^2)+(3x+6x)+(8-1)=16`
`\to 9x+7=16`
`\to 9x=9`
`\to x=1`
Vậy `x=1`
`b)`
`(x+2)(x^2-2x+4)-x(x^2-2)=15`
`\to x^3+2^3-x^3+2x=15`
`\to (x^3-x^3)+2x+8=15`
`\to 2x=7`
`\to x=7/2`
Vậy `x=7/2`
