I guess you could do this sort of quesion by raising the numbers to a power until they are whole numbers and then working it out.

Raising the first one by squaring it gives

\(6*9,100,49*2=54,100,98\)

therefore the result would be

\(3\sqrt{6},7\sqrt{2},10\)

for the second one squaring it would be

\(-4*3,-16,-9*2,-4*\frac{7}{2}=-12,-16,-18,-14\)

so it is

\(-3\sqrt{2},-4,-2\sqrt{\frac{7}{2}},-2\sqrt{3}\)

(note that I changed to negative because of that)

for the third one you need to cube it

\(21,27*2,21.952,8*5=21,54,21.952,40\)

Then it becomes

\(\sqrt[3]{21},2.8,2\sqrt[3]{5},3\sqrt[3]{2}\)

(Sorry for errors, I kind of rushed it)