Super cubes

Proposition: If $\sqrt[3]{a+\sqrt{b}}=x+\sqrt{y}$ ,then $\sqrt[3]{a-\sqrt{b}}=x-\sqrt{y}$

Proof : $a+\sqrt{b}=x^3+y\sqrt{y}+3x^2\sqrt{y}+3xy \\ \text{Equating rational,irrational parts,we get} \\ a=x^3+3xy,\sqrt{b}=3x^2\sqrt{y}+y\sqrt{y} \implies a-\sqrt{b}=x^3-3x^{2}\sqrt{y}+3xy-y\sqrt{y}=(x-\sqrt{y})^3 \\ \therefore \sqrt[3]{a-\sqrt{b}}=a-\sqrt{b}$

---

Now we have $\sqrt[3]{99-70\sqrt{2}}=x-\sqrt{y},\sqrt[3]{99+70\sqrt{2}}=x+\sqrt{y}.$ Multiply both the equations to get, $x^2-y=\sqrt[3]{99^2-70^2 \cdot 2}= \sqrt[3]{9801-9800}=\sqrt[3]{1}=1 \\ \therefore y=x^2-1 \ldots. (1)$ Now cube our original equation to get $99-70\sqrt{2}=x^3+3xy-y\sqrt{y}-3x^2 \sqrt{y}$ Equating rational parts we get $99=x^3+3xy$ .Then by equation $(1)$ ,we have $99=x^3+3x(x^2-1) \implies x(4x^2-3)=99$ By trial we find $x=3$ .By $x$ ,we get $y=\sqrt{8}$ . $\therefore \color{#20A900}{\boxed{x+y=11.}}$

Let $b=\sqrt[3]{99-70\sqrt{2}}$ and $a=\sqrt[3]{99+70\sqrt{2}}$ . Then, we are going to find $a+b$ and $ab$ .

$ab=\sqrt[3]{(99+70\sqrt{2})(99-70\sqrt{2})}=\sqrt[3]{9801-9800}=1$

Now let $x=a+b$ , we know that $x \in \mathbb{R}$ , so let's cube both sides:

$x^3=a^3+b^3+3ab(a+b)$

$x^3=99+70\sqrt{2}+99-70\sqrt{2}+3(1)x$

$x^3=3x+198 \implies x^3-3x-198=0$

By the rational root test we find that the only real root is $x=6$ , so $a+b=6$ .

Finally, we can find $a-b$ knowing that $a-b>0$ :

$a-b=\sqrt{(a+b)^2-4ab}=\sqrt{6^2-4(1)}=4\sqrt{2}$

Then, $b=\dfrac{(a+b)-(a-b)}{2}=\dfrac{6-4\sqrt{2}}{2}=\boxed{3-2\sqrt{2}}=3-\sqrt{8}$

Comparing we get $x=3$ and $y=8$ , so the final answer is $x+y=\boxed{11}$ .

We can try two thing 2 and 8 for y because it can't be any thing else