Question

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

Answer 1

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`.