Đáp án+Giải thích các bước giải:
`(x-1)^6-(x^2+1)^3+(2x)^3=0`
`<=>(x-1)^3(x-1)^3-(x^6+3x^4+3x^2+1)+8x^3=0`
`<=>(x^3-3x^2+3x-1)(x^3-3x^2+3x-1)-x^6-3x^4-3x^2-1+8x^3=0`
`<=>x^6-3x^5+3x^4-x^3-3x^5+9x^4-9x^3+3x^2+3x^4-9x^3+9x^2-3x-x^3+3x^2-3x+1-x^6-3x^4-3x^2-1+8x^3=0`
`<=>-6x^5+12x^4-12x^3+12x^2-6x=0`
`<=>-6(x^5-2x^4+2x^3-2x^2+x)=0`
`<=>x^5-x^4-x^4+x^3+x^3-x^2-x^2+x=0`
`<=>x^4(x-1)-x^3(x-1)+x^2(x-1)-x(x-1)=0`
`<=>(x-1)(x^4-x^3+x^2-x)=0`
`<=>(x-1)[x^3(x-1)+x(x-1)]=0`
`<=>x(x-1)(x-1)(x^2+1)=0`
`<=>x(x-1)^2(x^2+1)=0`
Vì `x^2>=0<=>x^2+1>=1>0`
`=>[(x=0),(x-1=0):}<=>[(x=0),(x=1):}`
Vậy `x\in{0;1}`
