#### Answer

(a)$$(f+g)(x)=f(x)+g(x)=(x^3+3x^2)+(3x^2-1) \\ =x^3+6x^2-1$$ $$D_{f+g}=\mathbb R$$
(b)$$(f-g)(x)=f(x)-g(x)=(x^3+3x^2)-(3x^2-1) \\ =x^3+1$$ $$D_{f-g}= \mathbb R$$
(c)$$(fg)(x)=f(x)g(x)=(x^3+3x^2)(3x^2-1) \\ =3x^5+9x^4-x^3-3x^2$$ $$D_{fg}=\mathbb R$$
(d)$$(f/g)(x)= f(x)/g(x)=\frac{x^3+3x^2}{3x^2-1}$$ $$D_{f/g}=\mathbb R - \{ \frac{1}{\sqrt{3}}, - \frac{1}{\sqrt{3}} \}$$

#### Work Step by Step

To find the domains, one should consider the following points:
The domain of $f+g$, $D_{f+g}$, is $D_f \cap D_g$ .
The domain of $f-g$, $D_{f-g}$, is $D_f \cap D_g$ .
The domain of $fg$, $D_{fg}$, is $D_f \cap D_g$ .
The domain of $f/g$, $D_{f/g}$, is $(D_f \cap D_g)- \{ \text{roots of } g(x) \}$ .
The domain of polynomial function is $\mathbb R$.