$a, P=(x+1)^3+(x+1)(6-x^2)-12$
$=x^3+3x^2+3x+1+6x-x^3+6-x^2-12$
$=(x^3-x^3)+(3x^2-x^2)+(3x+6x)+(6-12+1)$
$=2x^2+9x-5$
Vậy $P=2x^2+9x-5$
$b,$ Với $x=-\dfrac{1}{2}$ , ta có :
$P=2.\left (-\dfrac{1}{2} \right )^2+9.\left (-\dfrac{1}{2} \right )-5$
$=\dfrac{1}{2}-\dfrac{9}{2}-5$
$=-4-5$
$=-9$
Vậy $P=-9$ với $x=-\dfrac{1}{2}$
$c, P=0 ⇔ 2x^2+9x-5=0$
$⇔2x^2+10x-x-5=0$
$⇔2x(x+5)-(x+5)=0$
$⇔(2x-1)(x+5)=0$
$⇔\left[ \begin{array}{1}2x=1\\x=-5\end{array} \right.$
$⇔\left[ \begin{array}{1}x=\dfrac{1}{2}\\x=-5\end{array} \right.$
Vậy $x=\dfrac{1}{2}$ hoặc $x=-5$ thì $P=0$
