Which number in the monomial #125x^18y^3z^25# needs to be changed to make it a perfect cube?

1 Answer
Oct 19, 2016

Answer:

#z#'s exponent would need to be changed from #25# to a multiple of #3#

Explanation:

Using the properties #(x^a)^b = x^(ab)# and #(xy)^a = x^ay^a#, we can see that #(x^ay^b)^3 = (x^a)^3(y^b)^3 = x^(3a)y^(3b)#.

Notice that the #3# gets distributed to each of the exponents, meaning each of the exponents in the separated form has to be a multiple of #3#. Reversing these steps shows that a product of integers and variables is a perfect cube only if their powers are all multiples of #3#.

In this case:
#125 = 5^(3*1)#
#x^18 = x^(3*6)#
#y^3 = y^(3*1)#
#z^25 = z^(3*8+color(red)(1))#

Notice that the exponent of #z# is not a multiple of #3#, meaning that would need to be changed in order for the product to be a perfect cube. If it was, say, #24 = 3*8#, then we could write the product as

#125x^18y^3z^24 = 5^3x^(6*3)y^3z^(8*3)#

#=5^3(x^6)^3y^3(z^8)^3#

#=(5x^6yz^8)^3#

which is a perfect cube.