What is the square root of 125 times the cube root of 25?

##5^(13/6)##

Remember to find the prime factorisation of the numbers under the radicals, and that ##root(n){x} = x^{1/n}##

##sqrt{125}cdot root(3){25} = sqrt{5^{3}} cdot root(3){5^{2}} = (5^3)^{1/2) * (5^{2})^{1/3} = 5^(2/3)*5^(3/2) = 5^(2/3 + 3/2) = 5^(4/6 + 9/6) = 5^(13/6)##.

I'd argue this is as good as any form for simplicity, but it's the same as ##root(6){5^(13)}##. :)