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

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

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