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 :).

Let $$\alpha = \sqrt[3]{a} + \sqrt[3]{b} = 1 + \sqrt{49 + 20 \sqrt[3]{6}}$$
We can compute the minimal polynomial of the RHS:
(1 + sqrt(49 + 20 * (6)^(1/3))).minpoly()
x^3 - 72*x - 336

and then applying Cardano's formula we get $\alpha = 2\cdot (6^{1/3} + 6^{2/3})$.
We can transform this into $a,b$ form with $2^3 = 8$ and $6^2 = 36$:
$$\alpha = \sqrt[3]{48} + \sqrt[3]{288}$$
