the least value for :$\frac{a}{b^3+54}+\frac{b}{c^3+54}+\frac{c}{a^3+54}$

Let $\displaystyle g(x,y,z)=\frac{x}{y^3+54}+\frac{y}{z^3+54}+\frac{z}{x^3+54}$.

If we consider $g(a,b,1-(a+b))$ we can use $\partial_ag=0$ and $\partial_bg=0$ to numerically find the following critical points (up to cyclic permutation): $$\begin{array}{lll|l} \text{a} & \text{b} & \text{c} & \text{g(a,b,c)} \\ \hline 1.20836 & -0.608416 & 0.400057 & 0.0183912 \\ 0.51624 & 0.32016 & 0.1636 & 0.0185045 \\ 0.333333 & 0.333333 & 0.333333 & 0.0185058 \\ 4.69649 & -0.929554 & -2.76694 & 0.0424017\\ \end{array}$$ so in the region $a,b,1-(a+b)> 0$ it seems the minimum value is $\leq 0.0185045$. Now, for the boundaries, we use (w.l.o.g) $g(0,b,1-b)$. This has a minimum at $b=0.251454$ where $g(0,b,1-b)=0.0184826$. We also have to check So this seems to be the minimum value of $g(a,b,c)$ subject to $a+b+c=1$ and $a,b,c\geq 0$.

I tried to use Lagrange multipliers to get an exact algebraic solution, but the expressions were much too complicated (i.e. not solvable by radicals). I suspect there is no pretty closed form expression.

EDIT: By the way, here is a plot of $g(x,y,1-(x+y))$ in the region $x,y,1-(x+y)\geq 0$ (with critical points highlighted in blue and the minima highlighted in red):

Since you said non-negative, I am going to assume that zero is allowed in which case I found a minimum value of 0.0184826 achieved at $a=0, b=0.748545, c=0.251455$ up to their cyclic permutations.

After posting this I see that Oleg567 pointed to this solution already. Furthermore, if you want all $a,b,c$ to be strictly positive then it looks like the minimum is indeed at 0.0185045 although the values for $a,b,c$ where it is achieved are not unique. In addition to Alexander Gruber and daniel's answer, I found $a=0.530893, b=0.335498, c=0.133609$.