Is there some way to not use guess and check?

• Module 4 Day 6 Challenge Part 2 Mini-question

Is there some way to not use guess and check? Although it is a pretty good strategy, I find it quite unstable.

• @The-Blade-Dancer Hi again! This is a good question! Even though guess-and-check sometimes works and can even be the quickest approach sometimes, there's something a bit unsatisfying about it. How do we really know that we'll get anywhere? What if we end up guessing forever? $${\tiny \text{(The horror...) }}$$

Well, in this case, we can use a bit of algebra along with the fact that $$a$$ and $$b$$ are integers in order to figure out that $$a = 7$$ and $$b= 5.$$

Let's first start with the problem statement.

$$\sqrt{a} + \sqrt{b} = \sqrt[3]{22 \sqrt{7} + 26\sqrt{5}}$$

Cube both sides of the equation.

\begin{aligned} \left( \sqrt{a} + \sqrt{b} \right)^3 &= \left( \sqrt[3]{22 \sqrt{7} + 26\sqrt{5}} \right)^3 \\ a\sqrt{a} + 3a\sqrt{b} + 3b\sqrt{a} + b\sqrt{b} &= 22\sqrt{7} + 26\sqrt{5} \\ \left( 3b + a \right) \sqrt{a} + \left( 3a + b \right) \sqrt{b} &= 22\sqrt{7} + 26\sqrt{5} \\ \end{aligned}

That looks better..... right? (You don't agree it looks better?)

Now let's use the fact that $$a$$ and $$b$$ are integers. We know the right hand side has some radicals, the $$\sqrt{7}$$ and the $$\sqrt{5}$$ terms. Where could they have come from, though? Could either radical have come from the $$\left( 3b + a\right)$$ or the $$\left( 3a + b \right)$$? No, because $$a$$ and $$b$$ are both integers, so $$\left( 3b + a\right)$$ and $$\left( 3a + b \right)$$ are also both integers. Well, it looks like there's no doubt about where the $$\sqrt{5}$$ and the $$\sqrt{7}$$ came from.

$$\sqrt{5} \text{ and } \sqrt{7} \text{ came from the } \sqrt{a} \text{ and the } \sqrt{b}$$

So there you have it! Now that you know that $$a = 5$$ and $$b = 7$$ (or vice versa... it doesn't actually matter), then you can figure out that their sum is $$5 + 7 = \boxed{12}.$$