How do you simplify #root5(-t^10 )#?

1 Answer
George C.
Oct 6, 2015

#root(5)(-t^10) = -t^2#

Explanation:

If #a > 0# and #b, c >= 0#, then #a^(bc) = (a^b)^c#

Also #root(n)(-a) = -root(n)(a)# if #n in ZZ# is odd.

So #root(5)(-t^10) = -root(5)(t^10) = -(t^10)^(1/5) = -t^(10*1/5) = -t^2#

We can also see that #(-t^2)^5 = (-1)^5*(t^2)^5 = -t^10#, so #-t^2# is a fifth root of #-t^10#.

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