If I be the incenter of the triangle ABC and $x,y,z$ the circum radii of the triangles IBC,ICA & IAB, show that $4R^3-R(x^2+y^2+z^2)-xyz = 0$
$$\angle BID=90^{\circ}-\dfrac{B}{2}$$ $$\angle CID=90^{\circ}-\dfrac{C}{2}$$
$$\angle BIC=180^{\circ}-\dfrac{B+C}{2}$$ $$\angle BIC=180^{\circ}-\dfrac{180-A}{2}$$ $$\angle BIC=90^{\circ}+\dfrac{A}{2}$$
$$x=\dfrac{a}{2\sin\left(90^{\circ}+\dfrac{A}{2}\right)}$$ $$x=\dfrac{a}{2\cos\dfrac{A}{2}}$$
In the similar way:
$$y=\dfrac{b}{2\cos\dfrac{B}{2}}$$ $$z=\dfrac{c}{2\cos\dfrac{C}{2}}$$
$$R=\dfrac{abc}{4\triangle}$$
$$4R^3-R(x^2+y^2+z^2)-xyz$$ $$\dfrac{4(abc)^3}{64\triangle^3}-\dfrac{abc}{4\triangle}\left(\dfrac{a^2}{4\cos^2\dfrac{A}{2}}+\dfrac{b^2}{4\cos^2\dfrac{B}{2}}+\dfrac{c^2}{4\cos^2\dfrac{C}{2}}\right)-\dfrac{abc}{8\cos\dfrac{A}{2}\cos\dfrac{B}{2}\cos\dfrac{C}{2}}$$
$$\dfrac{1}{16}\left(\dfrac{(abc)^3}{\triangle^3}-\dfrac{abc}{\triangle}\left(\dfrac{a^2bc}{s(s-a)}+\dfrac{ab^2c}{s(s-b)}+\dfrac{abc^2}{s(s-c)}\right)-\dfrac{2(abc)^2}{s\triangle}\right)$$
$$\dfrac{(abc)^2}{16\triangle}\left(\dfrac{abc}{\triangle^2}-\dfrac{1}{s}\left(\dfrac{a}{s-a}+\dfrac{b}{s-b}+\dfrac{c}{s-c}\right)-\dfrac{2}{s}\right)$$
$$\dfrac{(abc)^2}{16\triangle}\left(\dfrac{abc}{\triangle^2}-\dfrac{1}{s}\left(\dfrac{a}{s-a}+\dfrac{b}{s-b}+\dfrac{c}{s-c}+2\right)\right)$$
Putting value of $s=\dfrac{a+b+c}{2}$
$$\dfrac{(abc)^2}{16\triangle}\left(\dfrac{abc}{\triangle^2}-\dfrac{2}{s}\left(\dfrac{a}{b+c-a}+\dfrac{b}{a+c-b}+\dfrac{c}{a+b-c}+1\right)\right)$$
$$\dfrac{(abc)^2}{16\triangle}\left(\dfrac{abc}{\triangle^2}-\dfrac{2}{s}\left(\dfrac{a(a+c-b)(a+b-c)+b(b+c-a)(b+a-c)+c(c+a-b)(c+b-a)+(a+c-b)(b+c-a)(a+b-c)}{(a+b-c)(b+c-a)(a+c-b)}\right)\right)$$
$$\dfrac{(abc)^2}{16\triangle}\left(\dfrac{abc}{\triangle^2}-\dfrac{2}{s}\left(\dfrac{a(a^2-b^2-c^2+2bc)+b(b^2-a^2-c^2+2ac)+c(c^2-a^2-b^2+2ab)+(b+c-a)(a^2-b^2-c^2+2bc)}{(a+b-c)(b+c-a)(a+c-b)}\right)\right)$$
$$\dfrac{(abc)^2}{16\triangle}\left(\dfrac{abc}{\triangle^2}-\dfrac{2}{s}\left(\dfrac{(b+c)(a^2-b^2-c^2+2bc)+b(b^2-a^2-c^2+2ac)+c(c^2-a^2-b^2+2ab)}{(a+b-c)(b+c-a)(a+c-b)}\right)\right)$$
$$\dfrac{(abc)^2}{16\triangle}\left(\dfrac{abc}{\triangle^2}-\dfrac{2}{s}\left(\dfrac{ba^2-b^3-bc^2+2b^2c+ca^2-cb^2-c^3+2bc^2+b(b^2-a^2-c^2+2ac)+c(c^2-a^2-b^2+2ab)}{(a+b-c)(b+c-a)(a+c-b)}\right)\right)$$
$$\dfrac{(abc)^2}{16\triangle}\left(\dfrac{abc}{\triangle^2}-\dfrac{2}{s}\left(\dfrac{ba^2-b^3+bc^2+b^2c+ca^2-c^3+b^3-ba^2-bc^2+2abc+c^3-a^2c-b^2c+2abc}{(a+b-c)(b+c-a)(a+c-b)}\right)\right)$$
$$\dfrac{(abc)^2}{16\triangle}\left(\dfrac{abc}{\triangle^2}-\dfrac{2}{s}\left(\dfrac{4abc}{(a+b-c)(b+c-a)(a+c-b)}\right)\right)$$
$$\dfrac{(abc)^2}{16\triangle}\left(\dfrac{abc}{\triangle^2}-\dfrac{abc}{s(s-a)(s-b)(s-c)}\right)$$
$$\dfrac{(abc)^2}{16\triangle}\left(\dfrac{abc}{\triangle^2}-\dfrac{abc}{\triangle^2}\right)$$
So this was a very long proof, in short time it is not feasible to do all of this, is there any shorter or smarter way to solve this question.