If I have a general d-dimension cubic form $C_0(x)$ with coefficients in $\mathbb{R}$ is it possible to find a cubic form $C_1(x)$ with coefficients in $\mathbb{R}$ such that for all $x\in\mathbb{R}^{d}$ the following are satisfied:
i) $C_1(x)=Q(x)L(x)$ where $Q(x)$ and $L(x)$ are quadratic and linear forms respectivly. ii)$\left|C_0(x)-C_1(x)\right|<\delta$ for some $\delta>0$.
Or is it possible that one can approximate $C_0(x)$ by $C_1(x)$ in some other way?
I would expect not. The reason being that if this is anything like true (products of quadratic forms and and linear forms being dense in the ordinary "strong" or norm topology - presumably what you mean) then more would be true than that. So for a proof one could either write down an equation that the coefficients of such products would satisfy (making them not Zariski dense). Or say something about the products being the image of an algebraic morphism, hence a constructible set. If this is even Zariski dense, it is huge.
Let $x = (x_1, ... x_d)$ and change coordinates so that $L(x) = x_1$. Since $C_0$ is irreducible it is, in particular, not divisible by $x_1$. Then, as mdeland says, staying on the hyperplane $x_1 = 0$, we see that $C_0$ is some nonzero form in the other variables which can get arbitrarily large.
