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Factorize $x^3-3$ over $\frac{\mathbb{Z}_7[y]}{\langle y^3-3 \rangle}$

It is clear that $y+\langle y^3-3 \rangle$ is a root of $x^3-3$. Then, I have $$x^3-3 = (x-y)(x^2+yx+y^2)$$ Can we factorize $x^2+yx+y^2$ further into linear factors?

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That quadratic's discriminant is ( all along we do arithmetic modulo $\;7\;$):

$$\Delta=y^2-4y^2=-3y^2=4y^2=(2y)^2$$

and thus its roots are

$$x_{1,2}=\frac{-y\pm2y}2=\begin{cases}\frac{-3y}2=4\cdot(-3y)=2y\\{}\\\frac y2=4y\end{cases}$$

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    why $\frac{1}{2}=4$? – 2017-02-03
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    @SHBaoS Because $\;2^{-1}=4\pmod 7\;$ , of course. Observe that Heptagon's answer is excellent and very short, though perhaps above the level you need as that's usually studied in Galois Theory. – 2017-02-03
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    $\Delta=4y^2=(5y)^2$ and that will also gives the same roots. – 2017-02-03
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    @SHBaoS Yes, of course, because $\;4=5^2=25\pmod 7\;$ , again... – 2017-02-03
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Yes, because finite extensions of finite fields are cyclic.

In particular, one can follow SpamIAm's comment and write $x^3-3=(x-y)(x-y^7)\left(x-y^{49}\right)$.

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    The multiplicative groups of finite fields are indeed cyclic, but I'm not clear how this applies to give the requested factorization. Please add at least one more line explaining that implication. – 2017-02-03
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    @hardmath The Galois group is cyclic and generated by the Frobenius map $\sigma: \alpha \mapsto \alpha^7$, so the roots of $x^3 - 3$ are $y, \sigma(y) = y^7$, and $\sigma^2(y) = y^{7^2}$. – 2017-02-04
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    @hardmath The OP does not seem to request the factorization, but rather asks whether his polynomial splits or not. – 2017-02-04
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    @SpamIAm Thanks! I added your point to the answer. – 2017-02-04
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Here's one last way to see that your polynomial splits completely. Just like over $\mathbb{C}$, the polynomial $x^3 - 3$ factors as $(x - \alpha)(x - \zeta \alpha)(x - \zeta^2 \alpha)$ in some algebraic closure, where $\zeta$ is a primitive $3^\text{rd}$ root of $1$ and $\alpha$ is any root. (Using your notation, $\alpha = y+\langle y^3-3 \rangle$).

However, in this case the base field $\mathbb{F}_7$ already contains $\zeta$: since the group of units $\mathbb{F}_7^\times$ is cyclic and $3 \mid 6 = \#\mathbb{F}_7^\times$, then $\mathbb{F}_7^\times$ contains an element of order $3$, i.e., a primitive $3^\text{rd}$ root of $1$. Indeed, since $2^3 = 8 = 1$, we see that $2$ is such a root. Thus the polynomial factors as $$ x^3 - 3 = (x - \alpha)(x - 2\alpha)(x - 4 \alpha) \, . $$