Cube Root

Given that:

$\begin{aligned} a+b\sqrt c & = \sqrt[3]{99-70\sqrt 2} & \small \color{#3D99F6} \text{Cubing both sides} \\ a^3+3a^2b\sqrt c + 3ab^2c + b^3c\sqrt c & = 99-70\sqrt 2 \end{aligned}$

Equating the rational and irrational parts:

$\begin{cases} a^3 + 3ab^2c = 99 & ...(1) \\ \left(3a^2b + b^3c\right)\sqrt c = -70\sqrt 2 & ...(2) \end{cases}$

It is reasonable to assume that $\sqrt c = \sqrt 2$ , $\implies c=2$ and the two equations become:

$\begin{cases} a\left(a^2 + 6b^2\right) = 99 & ...(1) \\ b\left(3a^2 + 2b^2\right) = -70 & ...(2) \end{cases}$

From $(1): a\left(a^2 + 6b^2\right) = 3^2 \times 11$ , since $a$ and $b$ are integers, $a$ can only be 1, 3, 9, 11, 33, and 99. Then we have:

$\begin{cases} a = 1 & \implies b = \sqrt {\dfrac {98}6} & \small \color{#D61F06} \text{No integral solution} \\ a = 3 & \implies b = \pm 2 & \small \color{#3D99F6} \text{Integral solution} \\ a = 9 & \implies \small \color{#D61F06} \text{No real solution} & \small \color{#D61F06} \text{No real solution for }a > 5 \end{cases}$

From $(2): b\left(3a^2 + 2b^2\right) = -70$ , we note that $b<0$ , since $3a^2 + 2b^2 > 0$ . Therefore, $b=-2$ . Substituting $a=3$ and $b=-2$ in $(2): -2\left(3(3^2) + 2(-2)^2\right) = - 70$ . Therefore, the solution $(a,b,c) = (3,-2,2)$ is valid and $a+b+c = \boxed{3}$ .