Inverse Tangent Identity

$\begin{aligned} \tan^{-1} \left(\frac AB\right) - \tan^{-1} \left(\frac 1C\right) & = \frac \pi 4 & \small \color{#3D99F6} \text{Given} \\ \implies \frac {\frac AB - \frac 1C}{1+\frac A{BC}} & = 1 \\ \frac {AC-B}{BC+A} & = 1 \\ AC - B & = BC+A \\ AC-BC & = A+B \\ AC-BC + C & = A+B+C & \small \color{#3D99F6} \text{Given that }A+B+C = 478 \\ \implies C(A-B+1) & = 478 = \color{#3D99F6} 2\times 239 & \small \color{#3D99F6} \text{Both 2 and 239 are primes.} \end{aligned}$

Since 2 and 239 are both primes this means that $C = 2$ or $C=239$ . From:

$\begin{aligned} \tan^{-1} \left(\frac AB\right) - \tan^{-1} \left(\frac 1C\right) & = \frac \pi 4 \\ \tan^{-1} \left(\frac AB\right) & = \frac \pi 4 + \tan^{-1} \left(\frac 1C\right) \\ \implies \frac AB & = \frac {1+\frac 1C}{1-\frac 1C} \end{aligned}$

$\implies \begin{cases} \text{If }C = 2, & \dfrac AB = \dfrac {1+\frac 12}{1-\frac 12} = \dfrac 31 & \implies A+B+C = 3+1+2= 6 \color{#D61F06} \ne 478 & \color{#D61F06} \text{Rejected} \\ \text{If }C = 239, & \dfrac AB = \dfrac {1+\frac 1{239}}{1-\frac 1{239}} = \dfrac {120}{119} & \implies A+B+C=120+119+239 \color{#3D99F6} = 478 & \color{#3D99F6} \text{Accepted} \end{cases}$

Therefore, $\sqrt[3]{A^3+B^3+C^3} = \sqrt[3]{120^3+119^3+239^3} \approx \boxed{257.456}$ .

The solution to this problem is on page 15 of this pdf file.