In Goldstein's Classical Mechanics, the cross product $\mathbf{V}^{*}$ of two vectors $\mathbf{D}$ and $\mathbf{F}$, with components $D_i$ and $F_i$ with respect to some Cartesian reference frame, respectively, is defined as a vector with components $$V_i^{*}=D_jF_k-D_kF_j,\qquad\text{$i,j,k$ in cyclic order.}$$ This definition implies that the direction of $\mathbf{V}^{*}$ depends on the handedness of the reference frame used, however, whilst in most elementary treatments of vector algebra, the cross product is defined as a vector with a well-defined direction independent of any reference frame, to be determined via the right-hand rule.
Indeed, when $\mathbf{D}=D\mathbf{i}$ and $\mathbf{F}=F\mathbf{j}$ with respect to a right-handed Cartesian reference frame, we have according to the above $$V_3^{*}=DF,\qquad V_1^{*}=V_2^{*}=0\qquad\implies\qquad\mathbf{V}^{*}=DF\mathbf{k}.$$ Now, the basis vectors $\mathbf{i}'=-\mathbf{i}$, $\mathbf{j}'=-\mathbf{j}$, $\mathbf{k}'=-\mathbf{k}$ and associated coordinate axes form a left-handed Cartesian reference frame. With respect to this reference frame, we have $\mathbf{D}=-D\mathbf{i}'$ and $\mathbf{F}=-F\mathbf{j}'$. Hence we now have $$V_3^{*}=(-D)(-F)=DF,\qquad V_1^{*}=V_2^{*}=0\qquad\implies\qquad\mathbf{V}^{*}=DF\mathbf{k}'=(-DF)\mathbf{k}.$$
I don't really understand the purpose of defining the cross product in a coordinate-dependent way which clearly yields different results depending on the handedness of the frame used. I understand that many physics texts wish to distinguish between two kinds of vectors (polar and axial), but in Goldstein's book this distinction merely seems to stem from trading the traditional, coordinate-independent geometrical definition for the above coordinate-dependent one.
The viewpoint of the above discussion is a passive one, i.e. the coordinate axes are inverted. Some texts prefer the active viewpoint, where instead of transitioning from one reference frame to another, points, vectors and physical objects are rotated with respect to one fixed reference frame. In this case one inverts all vectors: $\mathbf{D}'=-\mathbf{D}$ for all $\mathbf{D}$. If $\mathbf{V}^{*}=\mathbf{D}\times\mathbf{F}$, one would argue that $(\mathbf{V}^{*})'=\mathbf{D}'\times\mathbf{F}'=(-\mathbf{D})\times(-\mathbf{F})=\mathbf{D}\times\mathbf{F}=\mathbf{V}^{*}$. But why wouldn't $(\mathbf{V}^*)'=-\mathbf{V}^{*}$, i.e. why does one compute the cross product of the inverted vectors instead of inverting the cross product itself?