Radically Remarkable Root

$\implies 6x^3-3x^2-6x-4=0$

$6x^3=3x^2+6x+4$

Multiplying by $2$ and adding $\color{#3D99F6}{x^3}$ both sides.

$\implies \color{#3D99F6}{x^3}+12x^3=\color{#3D99F6}{x^3}+6x^2+12x+8$

$13x^3=(x+2)^3$

$\sqrt[3]{13}x=x+2$

Dividing by $x$ both sides.

$\implies x=2\left(\color{#20A900}{\dfrac{1}{\sqrt[3]{13}-1}}\right)$

$x=\dfrac{\sqrt[3]{13}+\sqrt[3]{169}+1}{6}$

$\therefore a+b+c=13+169+6=\boxed{188}$ .

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We know that: $a^3-b^3=(a-b)(a^2+ab+b^2)$

$\implies \dfrac{1}{a-b}=\dfrac{a^2+ab+b^2}{a^3-b^3}$

Putting $a=\sqrt[3]{13}$ and $b=1$ .

$\implies \color{#20A900}{\dfrac{1}{\sqrt[3]{13}-1}}=\dfrac{(\sqrt[3]{13})^2+\sqrt[3]{13}×1+(1)^2}{(\sqrt[3]{13})^3-(1)^3}=\dfrac{\sqrt[3]{13}+\sqrt[3]{169}+1}{12}$

We can work through the details which lead to Cardano's formula. "Completing the cube" makes the equation equal to $(6x-1)^3 - 39(6x-1) - 182 \; = \; 0$ Suppose that the roots of the cubic equation $X^3 - 39X - 182 = 0$ are given by $r = u + v \qquad \qquad s = \omega u + \omega^2v \qquad \qquad t = \omega^2u + \omega v$ where $\omega = \tfrac12(-1+i\sqrt{3})$ is the primitive cube root of $1$ . Then since $r + s + t \; = \; 0 \qquad \qquad rs + rt + st \; = \; -3uv \qquad \qquad rst \; =\; u^3 + v^3$ the roots can be expressed in this way provided that $u,v$ satisfy the equations $uv \; =\; 13 \qquad \qquad u^3 + v^3 \; = \; 182$ Thus $u^3$ and $v^3$ are roots of the quadratic $0 \; =\; X^2 - 182X +13^3 \; = \; (X -13)(X - 169)$ and it is clear that we can choose $u = \sqrt[3]{13}$ and $v = \sqrt[3]{169}$ , obtaining the three roots $\frac{\sqrt[3]{13} + \sqrt[3]{169} + 1}{6} \qquad \qquad \frac{\omega\sqrt[3]{13} + \omega^2\sqrt[3]{169} + 1}{6} \qquad \qquad \frac{\omega^2\sqrt[3]{13} + \omega\sqrt[3]{169} + 1}{6}$ of the original cubic. The real root is the first of these, and so the answer is $13 + 169 + 6 = \boxed{188}$ .