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)