Why is #(-2a^2b)^3sqrt(25a^4b^6)# equal to #-5/8# when a=1 and b=2, instead of #-5/128#?

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Answer 1 · 2018-02-14T01:50:06

See a solution process below:

Substitute #color(red)(1)# for each occurrence of #color(red)(a)#

Substitute #color(blue)(2)# for each occurrence of #color(blue)(b)#

#(-2color(red)(a)^2color(blue)(b))^3sqrt(25color(red)(a)^4color(blue)(b)^6)# becomes:

#(-2 xx color(red)(1)^2 xx color(blue)(2))^3sqrt(25 xx color(red)(1)^4 xx color(blue)(2)^6) =>#

#(-2 xx 1 xx 2)^3sqrt(25 xx 1 xx 64) =>#

#(-4)^3sqrt(1600) =>#

#-64 xx 40 =>#

#-2560#

Not sure where you are getting either #-5/8# or #-5/128#

Answer 2 · 2018-02-14T02:09:52

The answer is #-(5)/(8)# if the question is

#sqrt(25 a^4 b^6) -: ("-"2 a^2 b)^3#

I think that the question isn't printed correctly.

If the question is

#sqrt(25 a^4 b^6) -: (−2 a^2 b)^3#

then the answer is #-(5)/(8)#

#1)# Clear the parentheses in the denominator by raising all the factors inside to the power of #3#

#sqrt(25 a^4 b^6) -: ("-"2^3 a^6 b^3)#

#2)# Sub in #1# for #a# and #2# for #b#

#sqrt((25) (1^4) (2^6)) -: {("-"2^3) (1^6) (2^3)}#

#3)# Simplify by expanding the factors

#sqrt((5*5)(8*8))/ {("-"8)(8)}#

#4)# Find the square roots in the numerator

#( (5 ) (8)) / (("-"8) (8))#

#5)# Cancel the #8#s

#-(5)/(8)# #larr# answer