Question 116821
Your expressions have a little bit of ambiguity in them, but I think you mean:


1) {{{sqrt(4x^2-20x+25)}}} and
2) {{{root(3,-8n^12 (y^9))}}}



1) if you factor {{{4x^2-20x+25}}} you get {{{(2x-5)(2x-5)}}} (I'll leave it as an exercise for you to verify that is true - Hint: use FOIL).  Since {{{(2x-5)(2x-5)=(2x-5)^2}}}, we can say:


{{{sqrt(4x^2-20x+25)=sqrt((2x-5)^2)=2x-5}}} 


2) Again, let's factor the expression under the radical.  I'm not going to factor it completely, just enough to make it convenient to take the cube root.


 {{{-8n^12 (y^9)=(-2)(-2)(-2)(n^4)(n^4)(n^4)(y^3)(y^3)(y^3)}}}  (we can say this because we know that {{{(x^a)(x^b)=x^(a+b)}}})


So, we want {{{root(3,(-2)(-2)(-2)(n^4)(n^4)(n^4)(y^3)(y^3)(y^3))}}}


There are three factors of {{{-2}}}, so we take one of them,
There are three factors of {{{n^4}}}, so we take one of them,
and there are three factors of {{{y^3}}}, so we take one of them, and put them all together:


{{{root(3,-8n^12 (y^9))=-2n^4(y^3)}}}


And you can multiply {{{(-2n^4(y^3))(-2n^4(y^3))(-2n^4(y^3))}}} and verify that it equals {{{-8n^12 (y^9))}}} to check the answer.


Hope this helps
John