Question

How do you simplify `(12x^(2)y^(-2))^(5)(4xy^(-3))^(-8)`?

Answer 1

`(81x^2y^14)/(64)`

Very detailed answer. The more mathematically able can skip many parts.

Explanation:

Look at the two parts separately then put them together afterwards
and simplify. As a side note to aid understanding: if we were to have general form of `z^-2` this may be written as `1/(z^2)`

Consider the first part: `(12x^2y^(-2))^5`

Rewrite the bit inside the brackets as `(12x^2)/y^2`

So the whole part may be rewritten as:

`( (12x^2)/(y^2))^5`

A method example: `(x^2)^3` is the same as `x^2 times x^2 times x^2 =x^(2 times 3) = x^6`

Back to the question:

`( (12x^2)/(y^2))^5 =(12^5x^(2 times 5))/(y^(2 times 5)) = (12^5x^10)/y^10` ......... ( 1 )

Consider the second part: `(4xy^(-3))^(-8)`

Rewrite the bit inside the bracket as `(4x)/y^(3)`

So the whole part is may be written as:

`(4x/y^(3))^(-8)`

Although it is not normally written like this it gives:
`1/((4x)/(y^3))^8`

The net result of this is:

`((y^3)/(4x))^8 = (y^24)/(4^8x^8)` ....... ( 2 )

Putting (1) and (2) together gives:

`(12^5x^10)/y^10 times (y^24)/(4^8x^8)`

Grouping variables and constants we have:

( remember that `12^5 = (3 times 4)^5 = 3^5 times 4^5`)

so we have:

`(3^4 times 4^5)/4^8 times x^10/x^8 times (y^24)/(y^10)`

Simplifying -> `81/64 times x^2 times y^14`

`(81x^2y^14)/(64)`