knock knock... whose there? A Professor who doesn't teach.. happy halloween math

You know that sqrt(x) = x1/2 ?
You can do that and get: 3sqrt{20*a7*b6} = 3*{20*a7*b6}1/2 = 3*(20)1/2*(a7)1/2*(b6)1/2 = 3*201/2*(a7)1/2*(b6*(1/2))
As you see, only the member (b6)1/2 can be simplified and become b3. The other exponents can't be simplified, so we will turn them back into sqrt.
3*(b3)*201/2*(a7)1/2 = 3*(b3)*[20*(a7)]1/2 = 3*(b3)*sqrt[20*(a7)]
So A=3*(b3) and C=20*(a7)

Or, you can simply do: 3sqrt{20*a7*b6}=3sqrt{20*a7}*sqrt{b6} = 3*b3*sqrt{20*a7} . This is exactly the same method, but instead of using the exponent 1/2, we use sqrt.

It took me some time to figure it, but it's actually quite simple:
sqrt(7/8) = sqrt(7*2/8*2) = sqrt(14/16) = sqrt14/sqrt16 = sqrt14/4 = (1/4)*sqrt14
so A=1 B=4 C=14

Do you understand so far? I'll answer the rest later

Hey alexx! 

i'm following you on the sqrt7/8 problem but your losing me on 3sqrt{20 a^{7} b^{6}}

I tried plugging in your solution and it came out incorrect, so something must be afoot. hmm

Sorry for not updating, yes it seems that this solution can be further simplified.

Let's start again.
Let's mention some properties of the sqare root that we will use:
a) sqrt(a*b*c) = sqrt(a)*sqrt(b)*sqrt(c) (Note: the same doesn't apply for sqrt(a+b+c)
b) sqrt(ax) = ax/2 (as I said at the previous post, sqrtx= x1/2, and that property is the result of this.)
c) And, of course, the most basic property is sqrt(x)*sqrt(x)=x.

Now, let's start:
3sqrt(20*a7*b6) =(we use the property a)   3sqrt20*sqrt(a7)*sqrt(b6)
Now we will use the property b). As you see, it will only be useful on sqrt(b6) (if we use it on sqrt(20) or sqrt(a7) we won't get integer or mononym, since a mononym must have integer expotent)
And that's how we get to the result:
3sqrt(20)*sqrt(a7*b6/2 = 3*b3*sqrt(20)*sqrt(a7) = (using a again) 3*b3*sqrt(20*a7)
That's what I did before, but I had missed something. Let's go back to 3*b3*sqrt(20)*sqrt(a7). From there, we can do: 3*b3*sqrt(4*5)*sqrt(a7) = (using a) 3*b3*sqrt(4)*sqrt(5)*sqrt(a7) = 3*b3*2*sqrt(5)*sqrt(a7) = 6*b3**sqrt(5*a7)

Now, for the other exercises:
root4(x4*y2) (you know, sqrt means sqare root, which is the 2nd root only)
Anyways, the property a applies to all roots.
However, for the property b we have:
rooty(ax) = ax/y (in the case of the square root, y=2)
Similarly, for property c we have:
rooty(a)*rooty(ax)*.... (multiplying y times)... *rooty(ax) = a (so, for example root4(a)*root4(a)*root4(a)*root4(a)=a

So, we have: root4(x4*y2) = root4(x4) * root4(y2) = (using property b) x4/4 * y2/4 = x*y1/2
So, r=1 and s=1/2

Next:
(6*x2*y-4/5)5*(4*y2)1/2
Some properties that we will use:
d) (a*b*c)x = ax*bx*cx
e) (ax)y= ax*y
f) ax*ay = ax + y
And, of course, you probably know that a = a1
So, we have: (6*x2*y-4/5)5*(4*y2)1/2 = (using d)  65*(x2)5*(y-4/5)5*41/2*(y2)1/2
You know that 4=22, so we have:
65*(x2)5*(y-4/5)5*(22)1/2*(y2)1/2
Now, we use property e:
65*(x2*5)*(y(-4/5)*5)*(22*(1/2))*(y2*(1/2)) = 65*(x10)*(y-4)*2*y = 2*65*x10*y-4*y1 = (we use f for the y) 2*65*x10*y-4 + 1 = 2*65*x10*y-3

Now, 2*65=2*6*6*6*6*6=15552 (n=15552- use a calculator to be sure)
r=10
t=-3

Next:
These are just examples of property e. Should be easy:
(u3/2)4/5 = u(3/2)*(4/5) = u12/10 = u6/5 a= 6 b=5
(z4/3)9/2 = z(4/3)*(9/2) = z(36/6) = z6 a=6 b=1

Check for calculation mistakes
And, make 2 questions/thread other time

Ahhh yes most of this is clicking THANK YOU !!

For... Assume that a and b represent positive real numbers. Simplify the expression
3sqrt{20 a^{7} b^{6}}
into the simplest radical form A sqrt C}, where A and C are either integers or monomials.
Answer: A =  and C =

Still came up short for both a and c, but your logic makes sense, hmmm

The only other problem was for sqrt^4(x^4 y^2}
in which t= positive 3 not negative 3.

But really alexx thank you again!!

As an update I was able to figure out 3 sqrt(20 a^7 b^6)

alternative version of square root is 1/2

so root of 20
4      x        5
2 x 2

3 x 2=6

the tricky part was noting that instead of a^7, we can take an "a" out to make a^6

so,

(a^6)^1/2=
a^3

(b^6)^1/2=
b^3

ending results

6a^3b^3 sqrt(5a)

Sorry for not updating

That is correct, in addition about the other problem, I didn't see that it was xr/yt. Because it is on the denominator, the - will become +. (because a-x=1/ax)

It's perfectly okay,

Thank you again!