Let $a$ and $b$ be positive integers such that $(\sqrt[3]{a} + \sqrt[3]{b} - 1)^2 = 49 + 20 \sqrt[3]{6}.$ Find $a + b.$

Question

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 :).

Answer 1

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 $$(1+x\sqrt[3]{6}+y\sqrt[3]{6^2})^2=\\ 12xy+1+(x^2+2y)6^{2/3}+\sqrt[3]{6}(2x+6y^2)=\\ 49+20\sqrt[3]{6}$$ $$\begin{cases} 12xy+1=49\\ 2x+6y^2=20\\ x^2+2y=0 \end{cases}$$ $$\begin{cases} x=-2\\y=-2\end{cases}$$

Answer 2

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 $6^2 = 36$:

$$\alpha = \sqrt[3]{48} + \sqrt[3]{288}$$