In the triangle $\triangle ABC$, $|AB|^3 = |AC|^3 + |BC|^3$. Prove that $\angle ACB > 60^\circ$.

With your answer, to justify $\alpha \ge 120°$ means that you need to exclude the possibility of $\alpha = 60°$, which you should perhaps explicitly state. Also, keep in mind the comment by WW1 to somehow justify which side is largest.

Also, here is an alternative solution. With the equation of

$$\left\lvert AB \right\rvert^3 = \left\lvert AC \right\rvert^3 + \left\lvert BC \right\rvert^3 \tag{1}\label{eq1}$$

since all lengths are positive values, you automatically have that

$$\left\lvert AB \right\rvert \gt \left\lvert AC \right\rvert \tag{2}\label{eq2}$$ $$\left\lvert AB \right\rvert \gt \left\lvert BC \right\rvert \tag{3}\label{eq3}$$

Thus, $AB$ is the longest side. The largest angle is opposite the longest side (this is fairly generally known, but you can prove this using the sine law for acute angle triangles, and the cosine law for obtuse angles, such as shown in The largest angle in a triangle). If $\angle ACB \le 60°$, then the other $2$ angles are $\lt 60°$, giving a sum $\lt 180°$, which is not true. As such, $\angle ACB \gt 60°$.

Also, note this only depends on showing that $AB$ is the longest side. Any equation which provides this would also give the same conclusion.

In the standard notation we need to prove that $$\frac{a^2+b^2-c^2}{2ab}<\cos60^{\circ}$$ or $$c^2>a^2-ab+b^2$$ or $$\sqrt[3]{(a^3+b^3)^2}>a^2-ab+b^2$$ or $$(a+b)^2(a^2-ab+b^2)^2>(a^2-ab+b^2)^3$$ or $$ab>0,$$ which is true.

Id est, $$\measuredangle ACB>60^{\circ}$$ and we are done!