SOLUTION: {{{(5x^3y)/(x^2y^2)}}}.{{{y^3/(15x^2)}}} I remember going over cubes and I can solve the problems with {{{x^2}}} but the cubes confuse me.

Algebra ->  Rational-functions -> SOLUTION: {{{(5x^3y)/(x^2y^2)}}}.{{{y^3/(15x^2)}}} I remember going over cubes and I can solve the problems with {{{x^2}}} but the cubes confuse me.      Log On


   



Question 141325This question is from textbook McDougal Litell Algebra 2
: %285x%5E3y%29%2F%28x%5E2y%5E2%29.y%5E3%2F%2815x%5E2%29
I remember going over cubes and I can solve the problems with x%5E2 but the cubes confuse me.
This question is from textbook McDougal Litell Algebra 2

Answer by Edwin McCravy(20056) About Me  (Show Source):
You can put this solution on YOUR website!

Then until you get used to cubes and higher exponents
just get rid of all exponents first and write x%5E3
as x%2Ax%2Ax and y%5E2 as y%2Ay, etc.

Also break 15 into a product of primes
and write it as as 3%2A5

So,

%285x%5E3y%29%2F%28x%5E2y%5E2%29.y%5E3%2F%2815x%5E2%29

becomes

%285%2Ax%2Ax%2Ax%2Ay%29%2F%28x%2Ax%2Ay%2Ay%29.y%2Ay%2Ay%2F%283%2A5%2Ax%2Ax%29 

Indicate the multiplication of the numerators and
the denominators so that you only have one
fraction:

%285%2Ax%2Ax%2Ax%2Ay%2Ay%2Ay%2Ay%29%2F%283%2A5%2Ax%2Ax%2Ax%2Ax%2Ay%2Ay%29

Now cancel:



So all that's left is

%28y%2Ay%29%2F%283x%29

and we write y%2Ay as y%5E2 and end up with

y%5E2%2F%283x%29

This method will always work, and you can use it
until you can see that it's quicker just to subtract
the exponents, which would amount to subtracting
the number of x's or y's which would 
cancel in the mathod shown above.

Edwin