Let $a$ and $b$ be positive integers such that $(\sqrt[3]{a} + \sqrt[3]{b} - 1)^2 = 49 + 20 \sqrt[3]{6}.$ Find $a + b.$ Let $a$ and $b$ be positive integers such that $$(\sqrt[3]{a} + \sqrt[3]{b} - 1)^2 = 49 + 20 \sqrt[3]{6}.$$ Find $a + b.$
We let $\sqrt[3]a, \sqrt[3]b$ be $x,y,$ respectively. We expand LHS to get $$x^2+y^2+2xy-2x-2y+1=49+20\sqrt[3]6.$$ Now, I'm stuck. Help? Please share a clear solution :).

Okay. Hint: consider rational and irrational parts separately.
Solution:
Consider $(1+x\sqrt[3]{a}+y\sqrt[3]{b})^2$ for some rational $x,\,y$.
If $a$ or $b$ are other than $6$ and $6^2$ we would have $\sqrt[3]{a}$ and $\sqrt[3]{b}$ in the irrational part (as $(1+\color{blue}a+\color{red}b)^2=1+a^2+b^2+\color{blue}{2a}+\color{red}{2b}+2ab$), but we don't. So we search in the form
$$(1+x\sqrt[3]{6}+y\sqrt[3]{6^2})^2=\\
12xy+1+(x^2+2y)6^{2/3}+\sqrt[3]{6}(2x+6y^2)=\\
49+20\sqrt[3]{6}$$
$$\begin{cases}
12xy+1=49\\
2x+6y^2=20\\
x^2+2y=0
\end{cases}$$
$$\begin{cases}
x=-2\\y=-2\end{cases}$$

