How do you simplify #(20x^5)/y^2 * ((x^2y^2)/(2x))^3#?

1 Answer
KillerBunny
Oct 26, 2015

#5/2 x^8 y^4#

Explanation:

First of all, expand the power of the second factor: since the power of a fraction is the power of the numerator divided by the power of the denominator, we have that

#({x^2y^2}/{2x})^3=(x^2y^2)^3/(2x)^3#

Now, the power of a product is again the power of every single factor, so

#(x^2y^2)^3= (x^2)^3 * (y^2)^3#, and #(2x)^3 = 2^3 * x^3#.

The last rule we need is the one which states that when we deal with the power of a power, we must multiplicate the exponents:

# (x^2)^3 = x^{2*3)=x^6#, and the same goes for #(y^2)^3#

The result is thus

#({x^2y^2}/{2x})^3 = (x^6y^6)/(8x^3)#.

Now we're ready to multiply and cross simplify:

#(color(green)(20)x^5)/color(red)(y^2) * (color(blue)(x^6)color(red)(y^6))/(color(green)(8)color(blue)(x^3))#

So, we can simplify #20# and #8# dividing both by #4#, obtaining #5# and #2#.

Also, #y^2# cancels out and #y^6# becomes #y^4#, and #x^3# cancels out too and #x^6# becomes #x^3#.

So, we're left with

#5x^5 * (x^3y^4)/2= (5x^8 y^4)/2#

Related questions
See all questions in Exponential Properties Involving Products
Impact of this question
1494 views around the world
You can reuse this answer
Creative Commons License