Question

^5sqrt256x^7y^20

Answer 1

first notice that
\[256=2^8\] then do this in your head as follows.

Answer 2

5 goes in to 5 once with a remainder of 3 so
\[\sqrt[5]{2^8}=2\sqrt[5]{2^3}\]

Answer 3

how does 5 go into 5 once.... and has a remainder of 3.. 5/5=1....

Answer 4

5 goes in to 7 once with a remainder of 2, so
\[\sqrt[5]{x^7}=x\sqrt[5]{x^2}\]

Answer 5

damn i meant 5 goes in to 8 once with a remainder of 3, sorry

Answer 6

oooh lmao.. its okay i was just like...huh??

Answer 7

finally 5 goes in to 20 4 times exactly so
\[\sqrt[5]{y^{20}}=y^4\]

Answer 8

put this all together and get
\[\sqrt[5]{256x^7y^{20}}=2xy^4\sqrt[5]{2^3x^2}\] or if you prefer
\[\sqrt[5]{256x^7y^{20}}=2xy^4\sqrt[5]{8x^2}\]

Answer 9

yeah it is like that if you want to do it in your head. so for example
\[\sqrt[3]{x^{14}}=x^4\sqrt[3]{x^2}\] because 3 goes in to 14 4 times with a remainder of 2