Great question, @the-blade-dancer !

If you remember the binomial theorem at all from Module 3 (let me know if you need more explanation here), but we can use a shortcut for the expansion of \((\sqrt{a}+\sqrt{b})^3\).

In general, \((x+y)^3=x^3+3x^2y+3xy^2+y^3\), so

\((\sqrt{a}+\sqrt{b})^3=a\sqrt a +3a\sqrt b+ 3\sqrt a b+b\sqrt b\)
\((\sqrt{a}+\sqrt{b})^3= (a+3b)\sqrt a+(3a+b)\sqrt b\)

This gets us straight to the form that we need. And since this has to be equal to \(22\sqrt 7+26\sqrt 5\), it is most likely that \(a=7, b=5\), so we can go and check that out. After confirming that it is true, we get that our answer is \(7+5=\boxed{12}\). Let me know if you have any other questions!

@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 \( 5 + 7 = \boxed{12}.\)

🙂