How do i solve #root3{(ab^2c)/(64bc^2)} * root4{(16a^4)/(bc)# ?

#root3{(ab^2c)/(64bc^2)} * root4{(16a^4)/(bc)#

1 Answer
Jun 8, 2018

#(root(3)(a.b.c).a)/(2root(4)(b.c))#

Explanation:

We use exponential law:

#a^2 xx a^4# = #a^(2+4)# = #a^6#

#a^2/a^3# = #a^(2-3) = a^-1#

#1/a# = #a^-1#

#sqrta = a^(1/2)#

#root(3)a = a^(1/3)#

so we have:

#root(3)((ab^2c)/(64bc^2)) . root(4)((16a^4)/(bc))#

Step 1:

#root(3)((ab^2c)/(64bc^2))# = #(a^(1/3)b^(2/3)c)/(64^(1/3)b^(1/3)c^(2/3))#

#(a^(1/3)b^(2/3)c)/(64^(1/3)b^(1/3)c^(2/3))# = #((a^(1/3).b^(2/3 - 1/3).c^(1-2/3))/4)# = #color(red)((a^(1/3). b^(1/3).c^(1/3))/4)#

Step 2:

#root(4)((16a^4)/(bc))# = #(16^(1/4).a^(4/4))/(b^(1/4).c^(1/4)# = #color(red)((2.a)/(b^(1/4).c^(1/4))#

Step 3: Multiply and simplify the variables in color RED.

#color(red)((a^(1/3). b^(1/3).c^(1/3))/4)# . #color(red)((2.a)/(b^(1/4).c^(1/4))# = #(root(3)(a.b.c).a)/(2root(4)(b.c))#