Lời giải:
Đặt \(x-2=a(aeq 0)\). PT trở thành:
\(\left(\frac{a-1}{a}\right)^3-(a-1)^3=16\)
\(\Leftrightarrow 16+(a-1)^3-\left(1-\frac{1}{a}\right)^3=0\)
\(\Leftrightarrow 16+a^3-3a^2+3a-1-\left(1-\frac{3}{a}+\frac{3}{a^2}-\frac{1}{a^3}\right)=0\)
\(\Leftrightarrow a^3+\frac{1}{a^3}-3(a^2+\frac{1}{a^2})+3(a+\frac{1}{a})+14=0\)
\(\Leftrightarrow (a+\frac{1}{a})^3-3(a^2+\frac{1}{a^2})+14=0\)
\(\Leftrightarrow (a+\frac{1}{a})^3-3(a+\frac{1}{a})^2+20=0\)
Đặt \(a+\frac{1}{a}=t\Rightarrow t^3-3t^2+20=0\)
\(\Leftrightarrow t^2(t+2)-5(t^2-4)=0\)
\(\Leftrightarrow (t+2)(t^2-5t+10)=0\)
Dễ thấy \(t^2-5t+10>0, \forall t\in\mathbb{R}\Rightarrow t+2=0\Leftrightarrow t=-2\)
Do đó: \(a+\frac{1}{a}=-2\Leftrightarrow \frac{(a+1)^2}{a}=0\Rightarrow a=-1\)
\(\Rightarrow x=2+a=1\)
Vậy $x=1$ là nghiệm của pt.
