Đáp án+Giải thích các bước giải:
`(x+1)^3+(2x+1)^3=(3x+2)^3`
`<=> (x+1)^3-[(3x+2)^3-(2x+1)^3]=0`
`<=> (x+1)^3-(3x+2-2x-1)[(3x+2)^2+(3x+2)(2x+1)+(2x+1)^2]=0`
`<=> (x+1)^3-(x+1)[(3x+2)^2+(3x+2)(2x+1)+(2x+1)^2]=0`
`<=> (x+1)[(x+1)^2-(3x+2)^2-(3x+2)(2x+1)-(2x+1)^2]=0`
`<=> (x+1)[(x+1-3x-2)(x+1+3x+2)-(3x+2)(2x+1)-(2x+1)^2]=0`
`<=> (x+1)[(-2x-1)(4x+3)-(3x+2)(2x+1)-(2x+1)^2]=0`
`<=> (x+1)[(2x+1)(-4x-3)-(3x+2)(2x+1)-(2x+1)^2]=0`
`<=> (x+1)(2x+1)(-4x-3-3x-2-2x-1)=0`
`<=> (x+1)(2x+1)(-9x-6)=0`
`<=>`\(\left[ \begin{array}{l}x+1=0\\2x+1=0\\-9x-6=0\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x=-1\\x=\dfrac{-1}{2}\\x=\dfrac{-2}{3}\end{array} \right.\)
Vậy `x ∈ {-1;-1/2;-2/3}`
