How do you rationalize the denominator and simplify #root3((3a^8)/(4b))#?

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Answer 1 · 2017-05-19T16:25:06

#frac(root(3)(6 a^(8) b^(2)))(2 b)#

We have: #root(3)(frac(3 a^(8))(4 b))#

#= frac(root(3)(3 a^(8)))(root(3)(4 b))#

#= frac(root(3)(3 a^(8)))(root(3)(4 b)) cdot frac(root(3)(4 b))(root(3)(4 b)) cdot frac(root(3)(4 b))(root(3)(4 b))#

#= frac(root(3)(48 a^(8) b^(2)))(4 b)#

#= frac(root(3)(48) cdot root(3)(a^(8)) cdot root(3)(b^(2)))(4 b)#

#= frac(root(3)(2 cdot 2 cdot 2 cdot 6) cdot root(3)(a^(8)) cdot root(3)(b^(2)))(4 b)#

#= frac(root(3)(2 cdot 2 cdot 2) cdot root(3)(6) cdot root(3)(a^(8)) cdot root(3)(b^(2)))(4 b)#

#= frac(2 root(3)(6 a^(8) b^(2)))(4 b)#

#= frac(root(3)(6 a^(8) b^(2)))(2 b)#

Answer 2 · 2017-05-19T16:38:00

#color(blue)((root3(6a^8b^2))/(2b)#

#root3((3a^8)/(4b)#

#:.=((3a^8)/(4b))^(1/3)#

#:.=(3^(1/3)a^(8/3))/(4^(1/3)b^(1/3))#

#:.=(root3(3)root3(a^8))/(root3(4)root3(b))#

#:.=root3(3a^8)/(root3(4b))#

#:.=root3(3a^8)/(root3(4b)) xx (root3(4b))/(root3(4b))xx (root3(4b))/(root3(4b))#

#:.=root3(3a^8*4b*4b)/(4b)#

#:.=root3(4b) xx root3(4b) xxroot3(4b) =4b#

#:.=(root3(2*2*2*2*3a^8b^2))/(4b)#

#:.=root3(2) xx root3(2) xx root3(2)=2#

#:.=(cancel2^1root3(6a^8b^2))/(cancel4^2b)#

#:.=color(blue)(root3(6a^8b^2)/(2b)#