What are the irreducible components of $V(xy-z^3,xz-y^3)$ in $\mathbb{A}^3_K$?

What are the irreducible components of the algebraic set $V(xy-z^3,xz-y^3)$ in $\mathbb{A}^3_K$? (Here I'm letting $K$ be an algebraically closed field.)

Normally, what I do is take the equations determining an algebraic set $V(I)$, and usually one of them factors so that $V(I)$ decomposes as $V(J_1)\cup V(J_2)\cup\cdots$ or something. After breaking things down enough, I can eventually find that $K[x,y,z]/J_i$ is an intergral domain, so $J_i$ is prime, and $V(J_i)$ is irreducible.

However, I don't see any obvious way to break $V(xy-z^3,xz-y^3)$ down as a union of smaller sets since neither of the polynomials factor. I think $K[x,y,z]/(xy-z^3,xz-y^3)\cong K[x]\oplus \langle y,z\rangle$ because any occurrence of $xy$ or $xz$ be be replaced with a power of $z^3$ or $y^3$, but I'm not sure if that helps.

Let $(x,y,z) \in V(xy-z^3,xz-y^3)$. Two cases:

$1$. $x=0$. Then $y=z=0$.

$2$. $x\ne 0$. Then $y^4 = z^4$, so $z = \omega \cdot y$ where $\omega^4 = 1$. If $y=0$ then $z=0$. If $y \ne 0$ then $z \ne 0$ and we have $(x,y,z) = (\omega^3 y^2, y, \omega y)$.

Therefore we have one component, the line $y=z=0$, and for each $\omega$, a fourth root of $1$, we have a component isomorphic to $\mathbb{A}^1$: $\{ (\omega^3 t^2, t, \omega t )\ | \ t \in k \}$.