Giải thích các bước giải:
$(x-1)^3+x^3+(x+1)^3=(x+2)^3$
$⇔x^3-3x^2+3x-1+x^3+x^3+3x^2+3x+1=x^3+6x^2+12x+8$
$⇔(x^3+x^3+x^3)-(3x^2-3x^2)+(3x+3x)+(1-1)=x^3+6x^2+12x+8$
$⇔3x^3+6x-x^3-6x^2-12x-8=0$
$⇔2x^3-6x^2-6x-7=0$
$⇔-2(-x^3+4x^2-x+4x-x+4)=0$
$⇔-2[-x^2(x-4)-x(x-4)-(x-4)]$
$⇔-2(x-4)(-x^2-x-1)=0$
$⇔-2.[-(x-4)(x^2+x+1)]=0$
$⇔-(x-4)(x^2+x+1)=0$
$⇔(x-4)(x^2+x+1)=0$
$⇔$ \(\left[ \begin{array}{l}x-4=0\\x^2+x+1=0\end{array} \right.\) $⇔$ \(\left[ \begin{array}{l}x=4\\x∈∅\end{array} \right.\)
$\text{Vậy $x=4$}$
Học tốt!!!
