\(\displaystyle \sqrt{-x-2}-\sqrt[3]{x+5}<3\) In this inequation what i'd usually do is bring -x-2 and x+5 to the "same root" that's 6. But i don't think i'm going anywhere because of that 3 on the right. What would you suggest me to do here?

The condititon is \(\displaystyle x\in(-\infty,-2]\)

If you want to use the 6th roots

\(\displaystyle \sqrt{-x-2}=(-x-2)^{\frac{3}{6}}=\left(\sqrt[6]{-x-2}\right)^3\)

\(\displaystyle \sqrt[3]{x+5}=(x+5)^{\frac{2}{6}}=\left(\sqrt[6]{x+5}\right)^2\)

Instead...

\(\displaystyle x=-2\ \Rightarrow\ \sqrt{0}-\sqrt[3]{3}<3\)

\(\displaystyle x=-3\ \Rightarrow\ \sqrt{1}-\sqrt[3]{2}<3\)

\(\displaystyle x=-4\ \Rightarrow\ \sqrt{2}-\sqrt[3]{1}<3\)

\(\displaystyle x=-5\ \Rightarrow\ \sqrt{3}-\sqrt[3]{0}<3\)

\(\displaystyle x=-6\ \Rightarrow\ \sqrt{4}-\sqrt[3]{-1}=3\)