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

2 Answers
Jul 21, 2017

Answer:

#(((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)#

Jul 28, 2017

Answer:

#(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.