How do you simplify #sqrt(128c^6)#?

Question

1998 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-04-23T23:07:01

#sqrt(128c^6) = 8abs(c)^3sqrt(2)#

Note that if #x >= 0# then #sqrt(x)# denotes the non-negative square root of #x#.

So for any Real number #x# we find:

#sqrt(x^2) = sqrt(abs(x)^2) = abs(x)#

If at least one of #a, b >= 0# then:

#sqrt(ab) = sqrt(a)sqrt(b)#

So:

#sqrt(128c^6) = sqrt(2*64c^6) = sqrt(2*(8abs(c)^3)^2) = 8abs(c)^3sqrt(2)#