How do you simplify #\root3 { 125b ^ { 2} c ^ { 3} d ^ { 5} }#?

To simplify radical equations like his one, we an first find the cube root of the non-variable, #125#. The cube root of #125# is #5#, so let's move this outside the radical sign:

#5root(3)(b^2c^3d^5)#

Now let's work on the variables:

If the variable's exponent is less than the root (3), that variable cannot be simplified.

Since #b# is to the second power, the #b# can't be simplified further here, so we leave it inside

For the variables that can be simplified, the exponent of it outside the radical is equal to the exponent divided by the root (rounding down). The remainder that can't be divided is left inside.

With that being said, we simplify the #root(3)(c^3# as simply #c#, so we leave a #c# outside the radical:

#5croot(3)(b^2d^5)#

For #d#, we divide the exponent (5) by the root (3):

#5/3 = 1# #"remainder"# #2#

So the exponent of #d# outside the radical is #1#, and a #d^2# is left inside:

#color(red)(5cdroot(3)(b^2d^2)#

And this is simplified!

#root3(125b^2c^3d^5)#

#root3a xx root3a xx root3a=a#

#root3(5*5*5*b*b*c*c*c*d*d*d*d*d#

#5cdroot3(b^2d^2)#