How do you simplify #root5(2)/(3root5(162))#?

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Answer 1 · 2017-01-16T07:23:19

#3^-(9/5)#

Given, #root(5)2/[3xx root(5) 162]#

#rArr 2^(1/5)/[3xx(162)^(1/5)]#

#rArr 2^(1/5)/[3xx (2xx81)^(1/5)#

#rArr 2^(1/5)/[3xx2^(1/5)xx(3^4)^(1/5)#

#rArr cancel[2^(1/5)]/[cancel[2^(1/5)] xx3^1 xx 3^(4/5)#

#rArr 1/[3^(1+4/5)#

#rArr 1/[3^{(5+4)/5}]#

#rArr 1/3^(9/5)#

#rArr 3^-(9/5)#