How do you simplify #(\frac { 2x ^ { 7} y ^ { 8} } { 6x y ^ { 9} } ) ^ { 3}#?

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Answer 1 · 2017-04-22T12:13:48

See below.

1st you have to do everything inside the bracket and then cube it
#2x^7y^8# ÷ #6xy^9#
= #0.6x^6y^-1# × #0.6x^6y^-1# × #0.6x^6y^-1#
= #2x^18# #y^-3#

I believe that this is correct!
I hope this will help you!

Answer 2 · 2017-04-22T12:22:10

#color(red)((x^18)/(27y^3)#

#((2x^7y^8)/(6xy^9))^3#

#:.=(2^3x^21y^24)/(6^3x^3y^27)#

#:.=(cancel8^color(red)1x^(21-3))/(cancel216^color(red)27y^(27-24))#

#:.color(red)(=(x^18)/(27y^3)#

Answer 3 · 2017-04-22T15:13:25

#(x^18)/(27y^3)#

First consider the contents of the brackets.

#2/(2xx3) xx(x xxx^6)/x xx(y^8)/(y^8xxy)#

#2/2xx1/3xx x/x xxx^6 xxy^8/y^8 xx1/y#

#1xx1/3xx1xx x^6xx1xx1/y#

#(x^6)/(3y)#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Now put this back into the brackets

#((x^6)/(3y))^3 = (x^18)/(27y^3)#