Is it a unique triplet?

Let $x = 2+\sqrt[3]{3}$ . Therefore, $x-2 = \sqrt[3]{3}$ , $(x-2)^3 = 3= x^3-6x^2+12x-8$ .

$x^3-6x^2+12x = 11$

$x^2-6x = -12+ \frac{11}{x}$

$\large x = 6+\frac{-12+ \frac{11}{x}}{x}$

Therefore, $a = 6, b = 12, c = 11$ . Our answer is then $6+12+11 = \boxed{29}$ .

$\begin{aligned} \left(2 + \sqrt[3]{3}\right)^3 & = 8 + 12 \sqrt[3]{3} + 6 \sqrt[3]{3}^2 + 3 \\ & = 11 + 6\sqrt[3]{3} (2 + \sqrt[3]{3}) \quad \quad \small \color{#3D99F6}{\text{Dividing both sides by }2 + \sqrt[3]{3}} \\ \left(2 + \sqrt[3]{3}\right)^2 & = 6\sqrt[3]{3} + \frac{11}{2+\sqrt[3]{3}} \\ & = 6(2+\sqrt[3]{3}) - 12 + \frac{11}{2+\sqrt[3]{3}} \quad \quad \small \color{#3D99F6}{\text{Dividing both sides by }2 + \sqrt[3]{3}} \\ 2 + \sqrt[3]{3} & = 6 + \frac{ - 12 + \dfrac{11}{2+\sqrt[3]{3}}}{2+\sqrt[3]{3}} \end{aligned}$

$\implies a + b + c = 6+12+11 = \boxed{29}$

$2+\sqrt[3]{3} = a + \dfrac{-b+\frac{c}{2+\sqrt[3]{3}}}{2+\sqrt[3]{3}}\\ (2+\sqrt[3]{3})^2 = a(2+\sqrt[3]{3}) - b + \dfrac{c}{2+\sqrt[3]{3}}\\ (2+\sqrt[3]{3})^3 = a(2+\sqrt[3]{3})^2 -b(2+\sqrt[3]{3}) + c\\ 8+12\sqrt[3]{3} + 6\sqrt[3]{9} + 3 = a(4+4\sqrt[3]{3} + \sqrt[3]{9}) - 2b - \sqrt[3]{3}b + c\\ 11+12\sqrt[3]{3}+6\sqrt[3]{9} = 4a+ 4\sqrt[3]{3}a + \sqrt[3]{9}a - 2b - \sqrt[3]{3}b + c\\ 11+12\sqrt[3]{3}+6\sqrt[3]{9} = (4a-2b+c) + (4a-b)\sqrt[3]{3} + (a)\sqrt[3]{9}$

By comparison,

$a=6\\ 4a-b = 12 \implies 4(6) - b = 12 \implies b =24-12=12\\ 4a-2b+c=11 \implies 4(6)-2(12)+c = 11 \implies c=11$

Therefore, $a+b+c = 6+12+11 = \boxed{29}$

$2 +\sqrt[3]{3} = a+\frac{-b+\frac{c}{2+\sqrt[3]{3}}}{2+\sqrt[3]{3}}$

$(2-a)+\sqrt[3]{3} = \frac{-b+\frac{c}{2+\sqrt[3]{3}}}{2+\sqrt[3]{3}}$

$\left((2-a)+\sqrt[3]{3}\right)\left(2+\sqrt[3]{3}\right) = -b+\frac{c}{2+\sqrt[3]{3}}$

$(4-2a+b)+(4-a)\sqrt[3]{3}+\sqrt[3]{9} = \frac{c}{2+\sqrt[3]{3}}$

$\left((4-2a+b)+(4-a)\sqrt[3]{3}+\sqrt[3]{9}\right)\left(2+\sqrt[3]{3}\right) = c$

$(8-4a+2b)+(12-4a+b)\sqrt[3]{3}+(6-a)\sqrt[3]{9}+3 = c$

Now we know $c \in \mathrm{Z}$ , so $12-4a+b=0$ and $6-a=0$

This gives us that $a=6, b=12,$ and then plugging back in gives us $c=11$

Thus $a+b+c = \boxed{29}$

With $r= \sqrt[3]{3}$ and $x=2+r$ we can rewrite the equality as $x^3-ax^2+bx-c=0$ $11+12r+6r^2-a(4+4r+r^2)+b(2+r)-c=0$ $11-4a+2b-c +r (12-4a+b=0)+r^2 (6-a)=0$ Because $r$ and $r^2$ are irrational, this can only happen when

• $11-4a+2b-c=0$
• $12-4a+b=0$
• $6-a=0$

so that $a=6, b=12, c=11$ and the answer is $6+12+11=\boxed{29}$