How do you simplify #(13u²v(u^4v²w^7)³uw^6)/(-26(uv^4)³v³(uw²)²)#?

1 Answer
Jun 3, 2015

Starting with #(13u^2v(u^4v^2w^7)^3uw^6)/(-26(uv^4)^3v^3(uw^2)^2)#

it looks like it's going to be easiest to total up the powers of #u#, #v# and #w# separately in the numerator and denominator, using the properties of exponents like:

#x^ax^b = x^(a+b)#
#(x^a)^b = x^(ab)#

Numerator:
#u: 2+(4xx3)+1 = 15#
#v: 1+(2xx3) = 7#
#w: (7xx3)+6 = 27#

Denominator:
#u: (1xx3)+(1xx2) = 5#
#v: (4xx3)+3 = 15#
#w: (2xx2) = 4#

So the result of dividing numerator by denominator is:
#u: 15 - 5 = 10#
#v: 7 - 15 = -8#
#w: 27 - 4 = 23#

The scalar term is just #13/(-26) = -1/2#

Putting this all together we get:

#(13u^2v(u^4v^2w^7)^3uw^6)/(-26(uv^4)^3v^3(uw^2)^2)#

#= -(u^10w^23)/(2v^8)#

with restrictions #u != 0#, #v != 0#, #w != 0#.