3 root 2+4th root 64 +4th root 2500+6th root 8 is

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Solution

Dear Student,

$A s s u m i n g t h a t y o u r q u e s t i o n i s " 3 \sqrt{2} + \sqrt[4]{64} + \sqrt[4]{2500} + \sqrt[6]{8} = ? ", \sqrt[b]{a} c a n b e w r i t t e n a s a^{\frac{1}{b .}} a^{m} \times a^{n} = a^{m + n} B y P r i m e F a c t o r i s a t i o n, 64 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 2500 = 5 \times 5 \times 5 \times 5 \times 2 \times 2 8 = 2 \times 2 \times 2 3 \sqrt{2} + \sqrt[4]{64} + \sqrt[4]{2500} + \sqrt[6]{8} = 3 \sqrt{2} + {(64)}^{\frac{1}{4}} + {(2500)}^{\frac{1}{4}} + {(8)}^{\frac{1}{6}} = 3 \sqrt{2} + {(2^{6})}^{\frac{1}{4}} + {(5^{4} \times 2^{2})}^{\frac{1}{4}} + {(2^{3})}^{\frac{1}{6}} = 3 \sqrt{2} + (2^{\frac{3}{2}}) + (5 \times 2^{\frac{1}{2}}) + 2^{\frac{1}{2}} = 3 \sqrt{2} + (2^{\frac{2}{2}} \times 2^{\frac{1}{2}}) + (5 \sqrt{2}) + \sqrt{2} = 3 \sqrt{2} + 2 \sqrt{2} + 5 \sqrt{2} + \sqrt{2} = \sqrt{2} (3 + 2 + 5 + 1) = 11 \sqrt{2}$

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