​$$\text{Simplify: } \sqrt[6]{4096} + \sqrt[4]{50625} + \sqrt[3]{21952} + \sqrt{3364}$$​

Correct option is B

Given:
We need to simplify the expression:
​$$\sqrt[6]{4096} + \sqrt[4]{50625} + \sqrt[3]{21952} + \sqrt{3364}$$​​
Solution:
​$$4096 = 2^{12}\\\sqrt[6]{4096} = \sqrt[6]{2^{12}} = 2^{12/6} = 2^2 = 4\\50625 = 15^4\\\sqrt[4]{50625} = \sqrt[4]{15^4} = 15^{4/4} = 15\\21952 = 2^6 \times 7^3\\\sqrt[3]{21952} = \sqrt[3]{2^6 \times 7^3} = 2^{6/3} \times 7^{3/3} = 2^2 \times 7 = 4 \times 7 = 28\\3364 = 2^2 \times 29^2\\\sqrt{3364} = \sqrt{2^2 \times 29^2} = 2 \times 29 = 58\\$$​​
​$$\sqrt[6]{4096} + \sqrt[4]{50625} + \sqrt[3]{21952} + \sqrt{3364}$$​​
= 4 + 15 + 28 + 58 
= 105