Different ways to express $\sqrt{a}+\sqrt{b}$

You can use the conjugate:

\begin{align*} \sqrt{a}+\sqrt{b} &= \frac{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}\\\\ &=\frac{a^2-b^2}{\sqrt{a}-\sqrt{b}} \end{align*}

However, this does leave a difference of square roots in the denominator. In most cases, having a sum or difference of square roots in the numerator is less of a hassle than in the denominator.

Most of the time, you will use this technique of multiplying by the conjugate to remove pesky expressions in the denominator of a fraction.

---

It also works for complex numbers, for which multiplying by the conjugate can remove the imaginary part from a denominator:

\begin{align*} \frac{a+bi}{c+di}&=\frac{(a+bi)(c-di)}{(c+di)(c-di)}\\\\ &=\frac{(a+bi)(c-di)}{c^2+d^2} \end{align*}

---

Multiplying the entire expression by $\sqrt{ab}/\sqrt{ab}$ is interesting as well:

\begin{align*} \sqrt{a}+\sqrt{b} &= \frac{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{ab}\right)}{\sqrt{ab}}\\\\ &=\frac{a\sqrt{b}+b\sqrt{a}}{\sqrt{a}-\sqrt{b}} \end{align*}

---

If for some reason you were using a log table, this identity could be useful:

$$\sqrt{a}+\sqrt{b} = a^{1/2} + b^{1/2} = e^{(\ln a)/2} + e^{(\ln b)/2}$$

There is no nice general purpose "different way" to express that sum of roots.

I think what you are hoping for is something true to use where you wish that this was an equality $$ \sqrt{a + b} \ne \sqrt{a} + \sqrt{b} $$ but know better.

Sometimes multiplying both sides of an equation or both the numerator or denominator of a fraction by the conjugate $\sqrt{a} - \sqrt{b}$ will help.

Sometimes squaring both sides of an equation like $$ \sqrt{a} + \sqrt{b} = \text{something} $$ helps because after simplifying you have just the single square root $\sqrt{ab}$.