$\text{ The curve }y=a{x}^3+b{x}^2+cx+8\text{ touches }x\text{-axis at }\mathrm{P}(-2,0)\text{ and cuts the }$$\text{y-axis at a point Q where its gradient is 3. The values of a, b, c are respectively}$

$\begin{array}{l}\frac{dy}{dx}=3a{x}^2+2bx+c\\ \text{ Since the curve touches }\mathrm{x}\text{-axis at }(-2,0)\text{ so }\end{array}$

${\left.\frac{dy}{dx}\right|}_{(-2,0)}=0\Rightarrow 12a-4b+c=0----\left(i\right)$

The curve cut the y-axis at (0, 8)  so

${\left.\frac{dy}{dx}\right|}_{(0,8)}=3\Rightarrow c=3$

$\begin{array}{l}\text{ Also the curve passes through }(-2,0)\text{ so }\\ 0=-8a+4b-2c+8\Rightarrow -8a+4b-2=0----\left(ii\right)\end{array}$

$\begin{array}{r}\text{ Solving (i) and }\\ \text{ (ii) }a=-1/4,b=0\end{array}$