How do you simplify #8sqrt(11b^4) - sqrt(99b^4)#?

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Answer 1 · 2015-05-18T13:40:50

By exponential definition, we know that #root(n)(a^m)=a^(m/n)#

In this case, we can do the same for #b#, factoring it out:

#8b^2sqrt(11)-b^2sqrt(99)#
However, we can see that #99=11*9#, so

#8b^2sqrt(11)-b^2sqrt(11*9)#

However, #9# is a square number. Thus, we can take its root out of the square root.

#8b^2sqrt(11)-3b^2sqrt(11)#

However, as both factors are multiplying #b^2sqrt(11)#, we can factor it in function of such elements:

#b^2sqrt(11)(8-3)#