Question 89

The descending order of $$\sqrt[4]{10}, \sqrt[3]{6}, \sqrt{3}$$ is

Solution
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We raise all the terms to an equal power. Let's take the 12th power of each term. 
We get $$6^4,\ 10^3\ and\ 3^6$$

The value of the 4th power of 6 is easily over 1000, which is nothing but cube of 10. And lastly 6th power of 3 is 729.
Hence, the descending order of the terms would be the answer option. 

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