Something interesting in cubes

Go through the simplification below, $8x^{3}-12x^{2}-6x-1=0$ $16x^{3}-\left ( 8x^{3}+12x^{2} +6x+1\right )=0$ $16x^{3}-\left ( 2x+1 \right )^{3}=0$ $16x^{3}=\left ( 2x+1 \right )^{3}$ In considering in this way we may be able to find the real root with $a,b,c \in \mathbb{Z^{+}}$ satisfying given conditions, $\sqrt[3]{16}x=2x+1$ $x\left ( \sqrt[3]{16}-2\right )=1$ $x=\frac{1}{\sqrt[3]{16}-2}$ $x=\frac{\sqrt[3]{16^{2}}+2\sqrt[3]{16}+4}{8}$ $x=\frac{4\sqrt[3]{4}+4\sqrt[3]{2}+4}{8}$ $x=\frac{\sqrt[3]{4}+\sqrt[3]{2}+1}{2}\Rightarrow a+b+c=8$ $\therefore$ It concludes that the relavant answer is 8