Tangent bundle for the projective plane curve

Question

Consider the cubic $C$ with an equation $x_0^3+x_1^3+x_2^3=0$ (this is a projective curve on $\mathbb{P}_2=\mathbb{P}(V)$). I need to find the equation of the closure of all tangents to $C$ (it is called tangent bundle, isn't it?). I mean that each tangent line is a point of the dual space $\mathbb{P}(V^*)$.
So, first we note that each point of $C$ is smooth (because there is no $x\in C$ such that each partial derivative $\frac{\partial(x_0^3+x_1^3+x_2^3)}{\partial x_i}=3x_i^2$ is zero) hence we do not meet additional difficulties. A line on $\mathbb{P}_2$ is given by an equation $ax_0+bx_1+cx_2=0$ where $a,b,c$ are some numbers (elements of our field). We also know that each line must intersect each curve and our cubic in particular. Thus, we are to find come conditions for $a,b,c$ such that the corresponding line would be tangent to $C$.
To be honest it seems to be very complicated. However, could you help to do that?

Answer 1

No, this isn't the tangent bundle. :) We're talking about the dual curve $C^*\subset \Bbb P(V^*)$. Note that the tangent line to $C$ at a point $[z_0,z_1,z_2]$ has equation $$z_0^2(x_0-z_0)+z_1^2(x_1-z_1)+z_2^2(x_2-z_2) = z_0^2x_0+z_1^2x_1+z_2^2x_2=0,$$ so the corresponding point on the dual curve is $[z_0^2,z_1^2,z_2^2]$. (In general, Euler's Theorem on homogeneous functions tells us that when $f$ is homogeneous of degree $k$, we have $\nabla f(z)\cdot z = kf(z)$.)

The degree of $C^*$ will be the class of $C$, which, for a smooth curve of degree $k$, is $k(k-1)$. So you should be looking for an equation of degree $6$.