In $\triangle ABC$, $\left|\overline{AB}\right|=\left|\overline{AC}\right|$, $D$ is the foot of the perpendicular from $C$ to $AB$ and $E$ the foot of the perpendicular from $B$ to $AC$, then
$\textbf{(A)}~\left|\overline{BC}\right|^3>\left|\overline{BD}\right|^3+\left|\overline{BE}\right|^3$
$\textbf{(B)}~\left|\overline{BC}\right|^3<\left|\overline{BD}\right|^3+\left|\overline{BE}\right|^3$
$\textbf{(C)}~\left|\overline{BC}\right|^3=\left|\overline{BD}\right|^3+\left|\overline{BE}\right|^3$
$\textbf{(D)}~\text{None of the above}$
1. Given that $\triangle ABC$ is isosceles with $\left|\overline{AB}\right| = \left|\overline{AC}\right|$, we need to analyze the lengths of $\overline{BC}$, $\overline{BD}$, and $\overline{BE}$.
2. Let $AB = AC = x$ and $BC = y$. Since $D$ is the foot of the perpendicular from $C$ to $AB$, and $E$ is the foot of the perpendicular from $B$ to $AC$, we can use the properties of right triangles to find the lengths of $\overline{BD}$ and $\overline{BE}$.
3. In $\triangle ABC$, since $D$ and $E$ are the feet of the perpendiculars, $\triangle BDC$ and $\triangle BEC$ are right triangles.
4. Using the Pythagorean theorem in $\triangle BDC$ and $\triangle BEC$, we have:
   \[
   BD^2 + DC^2 = BC^2 \quad \text{and} \quad BE^2 + EC^2 = BC^2
   \]
5. Since $D$ and $E$ are the feet of the perpendiculars, $DC = EC = \frac{y}{2}$ (because $D$ and $E$ are midpoints of $BC$ in an isosceles triangle).
6. Therefore, we have:
   \[
   BD^2 + \left(\frac{y}{2}\right)^2 = y^2 \quad \text{and} \quad BE^2 + \left(\frac{y}{2}\right)^2 = y^2
   \]
7. Simplifying these equations, we get:
   \[
   BD^2 + \frac{y^2}{4} = y^2 \quad \Rightarrow \quad BD^2 = y^2 - \frac{y^2}{4} = \frac{3y^2}{4}
   \]
   \[
   BE^2 + \frac{y^2}{4} = y^2 \quad \Rightarrow \quad BE^2 = y^2 - \frac{y^2}{4} = \frac{3y^2}{4}
   \]
8. Therefore, $BD = BE = \frac{\sqrt{3}y}{2}$.
9. Now, we need to compare $\left|\overline{BC}\right|^3$ with $\left|\overline{BD}\right|^3 + \left|\overline{BE}\right|^3$:
   \[
   BC^3 = y^3
   \]
   \[
   BD^3 + BE^3 = \left(\frac{\sqrt{3}y}{2}\right)^3 + \left(\frac{\sqrt{3}y}{2}\right)^3 = 2 \left(\frac{\sqrt{3}y}{2}\right)^3 = 2 \cdot \frac{3\sqrt{3}y^3}{8} = \frac{3\sqrt{3}y^3}{4}
   \]
10. Comparing $y^3$ and $\frac{3\sqrt{3}y^3}{4}$:
    \[
    y^3 \quad \text{and} \quad \frac{3\sqrt{3}y^3}{4}
    \]
    Since $\frac{3\sqrt{3}}{4} \approx 1.299$, it is clear that:
    \[
    y^3 < \frac{3\sqrt{3}y^3}{4}
    \]
11. Therefore, $\left|\overline{BC}\right|^3 < \left|\overline{BD}\right|^3 + \left|\overline{BE}\right|^3$.

The final answer is $\boxed{\textbf{B}}$