Exercise 1.27 Harris Algebraic Geometry I'm somewhat perplexed by the hint given in Exercise 1.27. The exercise itself says the following:

Look at the map: $\mu: [X_0,X_1]\mapsto [X_0^3,X_0X_1^2,X_1^3]=[Z_0,Z_1,Z_2]$. Show that the image for this map is the cubic hypersurface given by the equation $Z_0Z_2^2=Z_1^3$.
Hint: Show that the line $\alpha Z_1-\beta Z_2$ meets the curve defined by $Z_0Z_2^2=Z_1^3$ exactly at $[1,0,0]$ and $\mu([\beta,\alpha])$.

Now, showing that the image of $\mu$ is contained in the cubic hypersurface and that it meets the line in the aforementioned points is an easy task. What I do not understand, however, is the following:
What good did the hint do me? How do I know now that the hypersurface is the image of $\mu$?
 A: Let me call the cubic hypersurface by $H$.
Now to see $H\subseteq \mu(\mathbb{C}P^1)$:
Let $P=[a_0, a_1, a_2]$ be a point on $H$. So that $a_0a_2^2=a_1^3$. If $a_0=0$ then $a_1=0$, so the point $P=[0: 0: 1]=\mu([0: 1])$.
So assume $a_0\neq 0$. So $P=[1: a_1: a_2]$ so that  $a_2^2=a_1^3$.
If further $a_2=0$, then $a_1=0$, so $P=[1:0: 0]=\mu([1: 0])$, we are done in that case.
So let us assume that $a_2\neq 0$.
Consider the line $a_2Z_1-a_1Z_2$ passing through the point $P$, by the Hint it meets $H$ at $[1:0:0]$ and $\mu[a_1, a_2]$.  Now check that $\mu[a_1, a_2]=P$.
To be honest, I haven't really (at least to my satisfaction) used the Hint in full force.
A: Question: "What good did the hint do me? How do I know now that the hypersurface is the image of μ?"
Answer: Another approach may be the following: If $\mathbb{P}^3$ has coordinates $x,y,z,w$ and if $f: \mathbb{P}^1 \rightarrow \mathbb{P}^3$ is the 3-uple embedding with
$$f(u:v):=(u^3:u^2v:uv^2:v^3)$$
it follows the ideal $I$ of $Im(f)$ is generated by the following polynomials:
$$I=(xz-y^2, yw-z^2, xw-yz).$$
It follows the polynomial $(xw-yz)w+(yw-z^2)z:=F(x,w,z)=xw^2-z^3 \in I$ is in the ideal. When you project onto $\mathbb{P}^2$ you get the map $g(u:v)=(u^3:uv^2:v^3)$ and $Im(g) \subseteq V(F)$. Maybe this helps in proving $Im(g)=V(F)$.
