1. \({3\sqrt{6}, 10, 7\sqrt{2}}\)
2.\({-2\sqrt{3}, -4, -3\sqrt{2}, -2\sqrt{\frac{7}{2}}}\)
3.\({\sqrt[3]{21}, 3 \sqrt [3]{2}, 2.8, 2 \sqrt[3]{5}}\)
They are trying to be tricky, here !!!
Remember that a√b is just √ [a^2 * b ]
And a = √a^2
And a∛ b is just ∛ [ a^3 * b ]
So
3√ 6 , 10 , 7√2 can be written as
√54 , √100 , √ 98 see ???.......you can arrange these for yourself
Similarly
-2√3 , -4 , -3√2, -2√ [7/2] becomes
- √12, -√16, -√18 and -√ [ (4 * 7) / 2 ] = - √14
Again you can arrange these....just remember that, for instance, -√13 > -√25 !!!!
Lastly
∛21 , 3∛2 , 2.8 , 2∛5 become
∛21, ∛54, ∛21.952 and ∛40
And you can handle this one, too !!!
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)