Đáp án:
$\text{- ÁP dụng hằng đẳng thức : (A+B)³ = A³ +3 . A².B + 3.A.B² +B³}$
$x^3 +(x+1)^3 =(1+2x)^3$
$⇔x^3 +x^3+3x^2+3x+1 =1 + 6x +12x^2+8x^3$
$⇔x^3+x^3-8x^3+3x^2-12x^2+3x-6x+1-1=0$
$⇔-6x^3-9x^2-3x=0$
$⇔ -6x^3-6x^2-3x^2-3x=0$
$⇔-6x^2(x+1)-3x(x+1)=0$
$⇔(x+1)(-6x^2-3x)=0$
$⇔(x+1)-3x(2x+1)=0$
$⇔$\(\left[ \begin{array}{l}x+1=0\\2x+1=0\\-3x=0\end{array} \right.\)
$⇔$\(\left[ \begin{array}{l}x=-1\\x=-\dfrac{1}{2}\\x=0\end{array} \right.\)
$\text{Vậy x ∈ {$-1 ; -\dfrac{1}{2}; 0 $ }}$
