To make it more straight forward lets convert the indices to fractional form.
#color(blue)("Introduction to the approach")#
By example:
Suppose we had #root(3)(2)# this is the same as #2^(1/3)#
Suppose we hade #root(3)(2^5)# this is the same as #2^(5/3)#
Suppose we hade #2^(1/3)xx2^(5/3)# this is the same as #2^(1/2+5/2) = 2^(6/3) = 2^2#
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#color(blue)("Answering the question")#
Given : # root(3)(x^2y) color(white)("d")( sqrt(xy)-root(5)(xy^3)color(white)("d"))#
#color(brown)("Using the above approach")#
Write as:
#color(red)(x^(2/3)y^(1/3))color(white)("d")(color(green)(color(white)("d")x^(1/2)y^(1/2) color(white)("ddd")-color(white)("ddd")x^(1/5)y^(3/5)color(white)("d")))#
#(x^(color(red)(2/3)color(green)(+1/2))xxy^(color(red)(1/3)color(green)(+1/2)))color(white)("d") - color(white)("d")( x^(color(red)(2/3)color(green)(+1/5))xxy^(color(red)(1/3)color(green)(+3/5)) )#
#color(white)("dd")(x^(7/6)xxy^(5/6)) color(white)("dd.d")-color(white)("ddd")(x^(13/15)xxy^(14/15) )" ".........Expression(1)#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Lets do a quick check on equivalents
Known that #1/2 = (1xx2)/(2xx2)=2/4#
So is the following also true: #3^(1/2) -> 3^(2/4)# ?
#sqrt(3)/root(4)(9) = 1# as required so we can do this.
Note that
#7/6 -> (7xx5)/(6xx5) =35/30#
#5/6->(5xx5)/(6xx5)=25/30#
#13/15->(13xx2)/(15xx2)=26/30#
#14/15->(14xx2)/(15xx2)=28/30#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
So #Expression(1)# becomes
#color(white)("dd")(x^(35/30)xxy^(25/30)) color(white)("dd.d")-color(white)("ddd")(x^(26/30)xxy^(28/30) )" ".........Expression(1_a)#
From this we can factor out the least values
#x^(26/30)y^(25/30)color(white)("d")(color(white)("d")x^(9/30)-y^(3/30))#
#x^(13/15)y^(5/6)( x^(3/10)-y^(1/10))#
Horrible numbers!!!!!!
