Cubes in a Triangle?

Since the question implies that the equation holds for any $AD \neq CD$ , consider the extreme case where $CD = 0$ .

Then $AD^3 - 0^3 = 25(AD - 0)$ which solves to $AD = 5$ , and $AD = AC = AB = BC = BD$ , so $BD = \boxed{5}$ .

Let $BD=x$ , the side length of $\triangle ABC$ be $a$ , $AD=b$ , and $CD=c$ . Then $b+c=a$ and

$\begin{aligned} b^3-c^3 & = 25(b-c) \\ (b-d)(b^2 + bc + c^2) & = 25(b-c) \\ b^2 + bc + c^2 & = 25 \end{aligned}$

Using cosine rule , we have:

$\begin{cases} BD^2 = AB^2 + AD^2 - 2AB \cdot AD \cos 60^\circ & \implies x^2 = a^2 + b^2 - ab &...(1) \\ BD^2 = BC^2 + CD^2 - 2BC \cdot CD \cos 60^\circ & \implies x^2 = a^2 + c^2 - ac &...(2) \end{cases}$

$\begin{aligned} (1)+(2): \quad 2x^2 & = 2a^2 + b^2 + c^2 - a(b+c) \\ & = 2a^2 + b^2 + c^2 - a^2 \\ & = a^2 + b^2 + c^2 \\ & = (b+c)^2 + b^2 + c^2 \\ & = b^2 + 2bc + c^2 + b^2 + c^2 \\ & = 2(b^2 + bc + 2c^2) \\ \implies x^2 & = b^2 + bc + 2c^2 = 25 \\ x & = \boxed 5 & \small \color{#3D99F6} \text{Since }x > 0 \end{aligned}$