Giải thích các bước giải:
Ta có:
(x³-8) - (x-2)(x²+8x-20) = 0
⇔(x-2)(x²+2x+4) - (x-2)(x²+8x-20) = 0
⇔(x-2)[(x²+2x+4) - (x²+8x-20)] = 0
⇔(x-2)(x²+2x+4-x²-8x+20) = 0
⇔(x-2)(-6x+24) = 0
⇔(x-2)×(-6)×(x-4)=0
⇔(x-2)(x-4)=0
⇔\(\left[ \begin{array}{l}x-2=0\\x-4=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=2\\x=4\end{array} \right.\)