$x³-x²-21x+45=0
$⇔x³-3x²+2x²-6x+15x+45=0$
$⇔x²(2-3)+2x(x-3)-15(x-3)=0$
$⇔(x-3)(x²+2x-15)=0$
$⇔(x-3)(x²-3x+5x-15)=0$
$⇔(x-3)[x(x-3)+5(x-3)]=0$
$⇔(x-3)²(x+5)=0$
$⇔$ \(\left[ \begin{array}{l}(x-3)²=0\\x+5=0\end{array} \right.\) ⇔\(\left[ \begin{array}{l}x-3=0\\x=-5\end{array} \right.\) ⇔\(\left[ \begin{array}{l}x=3\\x=-5\end{array} \right.\)
$vậy$ $x=3$ $hoặc$ $x=-5$