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Numbers are to be  arranged in ascending order. 

\sqrt[8]{90},\ \ \   \sqrt[4]{10},\ \ \ \sqrt6

We have 8th root of 90.  So we take 6 raised to 4 and 10 raised to 2, to start with along with 90.

we take an initial set :

we\ know\ 81<90<100<36*36\\\\\sqrt{81}\ \textless \ \sqrt{90}\ \textless \ \sqrt{100}\ \textless \ \sqrt{36*36}\\\\So\ \ 9\ \textless \ \sqrt{90}\ \textless \ 10\ \textless \ 36\\\\3\ \textless \ \sqrt[4]{90}\ \textless \ \sqrt{10}\ \textless \ 6\\\\\sqrt3\ \textless \ \sqrt[8]{90}\ \textless \ \sqrt[4]{10}\ \textless \ \sqrt6

The ascending order is clear from the above relation.
3 3 3
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you could also solve the exercise by squaring the given quantities 3 times. Then you will get 90, 100 and 6^4. Then we know which is greater.