(x+1)(x+2)(x+5)-$x^{2}$. (x+8)=27
⇔($x^{2}$ +3x+2)(x+5)-($x^{3}$ +8$x^{2}$ )=27
⇔$x^{3}$ +5$x^{2}$ +3$x^{2}$ +15x+2x+10-$x^{3}$ -8$x^{2}$ =27
⇔($x^{3}$ -$x^{3}$ )+(5$x^{2}$ +3$x^{2}$ -8$x^{2}$ )+(15x+2x)+10=27
⇔17x+10=27
⇔17x=27-10
⇔17x=17
⇔x=17:17
⇔x=1.
Vậy x=1.