How do you simplify #((2x^3y^2) /( 3xy))^ -3#?

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Answer 1 · 2016-03-25T01:33:24

#27/(8x^3y^2)#

The equation is raised to the negative third power, flip the fraction to turn it to a positive third power:

#((2x^3y^2)/(3xy))^-3 = ((3xy)/(2x^3y^2))^3#

Then raise the numerator and the denominator by the third power:

#(3xy)^3/(2x^3y^2)^3#

Distribute the third power exponent:

#(3^3x^3y^3)/(2^3x^9y^6)#

Factor an #x^3y^3# from top and bottom:

#((x^3y^3)(3^3))/((x^3y^3)(2^3x^3y^2))#

Cancel out like terms from top and bottom:

#(3^3)/(2^3x^3y^2)#

Simplify: #27/(8x^3y^2)#

Answer 2 · 2016-03-26T22:16:15

#27/(8x^6y^3)#

For a moment let us disregard the power outside the brackets.

Example:#" " 1/x^2 # can be written as #" "x^(-2)#

So we have#" " 2/3xxx^3y^2xxx^(-1)y^(-1)#

This gives us:#" "2/3xx x^(3-1)y^(2-1)#

#2/3xxx^2y = (2x^2y)/3#

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Example: Suppose you have#" " (1/x)^(-3)# then this is#" "(x/1)^3#

Ok, so your question has: #" "((2x^2y)/3)^(-3)#

This gives us:#" "(3/(2x^2y))^3#

#3^3= 27#

#2^3=8#

#(x^2)^3 =x^(2xx3) = x^6#

#(y)^3 = y^3#

Putting it all together

#27/(8x^6y^3)#