Let us consider the splitting field $L$ of $f = x^3 - 135x - 270$ over $\mathbb{Q}_5$, and let $\alpha_1,\alpha_2,\alpha_3 \in L$. Let $\beta = (\alpha_2 - \alpha_1)^3$.
Question: How do I check if $L$ contains a square root of $\beta$?
I tried to approach this problem by noting that $ L = \{ c_0 + c_1 \alpha + c_2 \alpha_2^2 \: | \: c_i \in \mathbb{Q}_5 \} $ for any $\alpha = \alpha_1,\alpha_2,\alpha_3$.
For $\gamma = c_0 + c_1 \alpha + c_2 \alpha_2^2$ we have
$$ \begin{align} \gamma^2 &= c_0^2 + 2 c_0 c_1 \alpha + (2 c_0 c_2 + c_1^2) \alpha^2 + 2 c_1 c_2 \alpha^3 + c_2^2 \alpha^4 \\ & = c_0^2 + 2c_0 c_1 \alpha + (2 c_0 c_2 + c_1^2) \alpha^2 + 2 c_1 c_2 (135 \alpha + 270) + c_2^2 (135 \alpha^2 + 270 \alpha) \\ &= c_0^2 + 540 c_1 c_2 + (2 c_0 c_1 + 270 c_1 c_2 + 270 c_2^2) \alpha + (2 c_0 c_2 + c_1^2 + 135 c_2^2) \alpha^2. \end{align} $$
Also, we have $$ \begin{align} \beta = (\alpha_2 - \alpha_1)^2 &= \alpha_2^3 - 3 \alpha_1 \alpha_2^2 + 3 \alpha_1^2 \alpha_2 - \alpha_1^3 \\ &= -135 \alpha_1 + 135 \alpha_2 + 3 \alpha_1^2 \alpha_2 - 3 \alpha_1 \alpha_2^2 \end{align} $$
Since I want to have $\gamma^2 = \beta$, I could compare the coefficients of both terms for choosing $\alpha$ as $\alpha_1$, $\alpha_2$ or $\alpha_3$. But this approach does not seem to be very promising since I would receive an armada of equations over $\mathbb{Q}_5$ with quadratic variables.
Is there any other approach for this problem? Maybe with valuations or something similar?