Question

How do you simplify `(\frac { ( 4s ^ { 3} t ^ { - 2} ) ( 2t - 3u ^ { 5} ) } { 8u ^ { 7} } ) ^ { 2}`?

Answer 1

`(((4s^3t^(-2))(2t-3u^5))/(8u^7))^2=(s^6(2t-3u^5)^2)/(4t^4u^14)`

Explanation:

`(((4s^3t^(-2))(2t-3u^5))/(8u^7))^2`

= `((4^2(s^3)^2(1/t^2)^2(2t-3u^5)^2)/(8^2(u^7)^2))`

= `(16s^(3xx2)(1^2/t^(2xx2))(2t-3u^5)^2)/(64u^(7xx2))`

= `(cancel16s^6(1/t^4)(2t-3u^5)^2)/(cancel16xx4u^14)`

= `(s^6(2t-3u^5)^2)/(4t^4u^14)`

Answer 2

`(s^6(2t-3u^5)^2)/(4t^4u^14)`

Explanation:

`(((4s^3t^color(blue)(-2))(2t-3u^5))/(8u^7))^2`

Simplify inside the bracket first:

`= ((cancel4s^3(2t-3u^5))/(cancel8^2t^color(blue)(2)u^7))^2`

Square each factor in the bracket:

`(s^6(2t-3u^5)^2)/(2^2t^4u^14)`

We could also square the binomial but there seems little point.
We would not be able to cancel any factors (there will be + and -signs.
There will not be like terms nor a common factor
The answer is already in factor form.