Question

What is `((3^-1a^4b^-3)^-2)/((6a^2b^-1c^-2)^2)`?

Answer 1

`(b^8c^4)/(4a^12)`

Explanation:

There are lots of different ways to do this, but these are the steps I followed:

Using the index law `(a^m)^n=a^(mn)` you can simplify as follows:

`(3^(-1*(-2))a^(4*(-2))b^((-3)*(-2)))/(6^(1*2)a^(2*2)b^(-1*2)c^(-2*2))=(3^(2)a^(-8)b^6)/(6^2a^4b^(-2)c^(-4))`

Using the index law `a^m/a^n=a^(m-n)`, you can remove the values of `a` and `b` from the denominator (the bottom of the fraction), giving

`(3^2a^(-8-4)b^(6-(-2)))/(6^2c^(-4))=(3^2a^(-12)b^8)/(6^2c^(-4))`

Using the index law `a^(-n)=1/a^n`, and conversely `1/a^(-n)=a^n`, the next step would be to swap around the values so that they all have positive indices:

`(3^2b^8c^4)/(6^2a^12)`

Simplifying gives:

`(3^2b^8c^4)/(6^2a^12)=(9b^8c^4)/(36a^12)=(b^8c^4)/(4a^12)`

Answer 2

`(b^8c^4)/(4a^12)`

Explanation:

`((3^-1a^4b^-3)^-2)/(6a^2b^-1c^-2)^2`

`:.color(red)((a^m)^n=a^(mn)`

`:.=(3^(color(red)(-1 xx -2))a^color(red)((4 xx -2))b^color(red)(-3 xx -2))/(6^(color(red)color(red)(1 xx 2))a^color(red)(2 xx 2)b^(color(red)(-1 xx 2))c^(color(red)(-2 xx 2))`

`:.=(3^2a^-8b^6)/(6^2a^4b^-2c^-4)`

`:.=(9a^-8b^6)/(36a^4b^-2c^-4)`

`:.=(9/1 xx 1/a^8 xx b^6/1)/((36a^4)/1 xx 1/b^2 xx 1/c^4)`

`:.=((9b^6)/a^8)/((36a^4)/(b^2c^4))`

`:.=color(red)(a^m xx a^n=a^(m+n)`

`:.=(9b^6)/(a^8) xx (b^2c^4)/(36a^4)`

`:.=(cancel9^color(red)1b^8c^4)/(cancel36^color(red)4a^12)`

`:.=(b^8c^4)/(4a^12)`