Think algebra would be easy?

First of all,

$\displaystyle \begin{aligned} a&=\sqrt{b^2 + c^2 - 2bc\cos A}\\s& = \dfrac{a+b+c}{2}~~~~\text{NOTE: }s\text{ is the semi~perimeter of the } \bigtriangleup\\\\ \Delta &= \frac{1}{2}bc\sin A ~~~~\text{NOTE: }\Delta \text{ is the area of the } \bigtriangleup\\ r_1 & = \dfrac{\Delta}{s-a}~~\text {which is ex-radius}_1 \\ r_2 & = \dfrac{\Delta}{s-b}~~\text {which is ex-radius}_2 \\ r_3 & = \dfrac{\Delta}{s-c}~~\text {which is ex-radius}_3 \end{aligned}$

Now,

$\begin{aligned} r_1 + r_2 + r_3 - r &= (r_1 - r) + (r_2 + r_3) \\&= 4R\sin^2 \frac{A}{2} + 4R \cos^2 \frac{A}{2}\\&=4R\\\\ \Rightarrow r_1 + r_2 + r_3 &= 4R + r\\\\ \text{And, } r_1r_2 + r_2r_3 + r_3r_1 &= \Delta^2\left(\frac{1}{(s-a)(s-b)} + \frac{1}{(s-b)(s-c)} + \frac{1}{(s-c)(s-a)}\right)\\ &=\Delta^2 \times\frac{s^2}{\Delta^2} \\ &= s^2\\\\ \text{And, } r_1r_2r_3 &= \frac{\Delta^3 \times s}{\Delta^2}\\ &=\Delta s \\&= rs^2\\\end{aligned}$

So, the roots of the equation $x^3 - (4R+r)x^2 + s^2x - s^2r=0$ becomes $r_1, ~r_2$ and $r_3$

Solving,

$\begin{aligned} a &= \sqrt{b^2 + c^2 - bc}\\&=\sqrt{65 - 28} \\&=\sqrt{37}\\ s &= \dfrac{11 + \sqrt{37}}{2}\\\\ \Delta = \frac{1}{2}bc \sin A &= 14\cdot\frac{\sqrt{3}}{2}\\&=7\sqrt{3}\\\\ x_1 = r_1 &= \frac{\Delta}{s-a} \\&=\frac{14\sqrt{3}}{11-\sqrt{37}}\\\\ x_2 = r_2 &= \frac{\Delta}{s-b} \\&=\frac{14\sqrt{3}}{\sqrt{37} - 3}\\\\ x_3 = r_3 &= \frac{\Delta}{s-c} \\&=\frac{14\sqrt{3}}{\sqrt{37} + 3}\\\\ \therefore x_1^3 + x_2^3 + x_3^3 &\approx \boxed{625.634}\end{aligned}$