xb^0y(x^-2b^-2)^2/xb(by^0)xby?

Answer 1

First I assume this is one fraction with a product on top and a product on bottom:
xb^0y(x^-2b^-2)^2 / xb(by^0)xby

Anything to the power 0 is 1, so replace and remove those:
x(1)y(x^-2b^-2)^2 / xb(b(1))xby
xy(x^-2b^-2)^2 / xbbxby

Distribute the square in the numerator by multiplying the exponents.
xy(x^-4b^-4) / xbbxby

Now you can switch those negative powers and put them in the denominator instead:
xy / [ xbbxby x^4b^4 ]

Now collect your like terms in the denominator:
xy / ( x^6 b^7 y )

Cancel one x and one y:
1 / ( x^5 b^7)

So the answer is:
1 / (x^5 b^7)

Answer 2

As written: X. The key here is the first raised power (xb raised to 0y). Zero times anything is zero. Zero raised to any positive power is zero. Any number raised to the zero power (in this case, b) is 1. 1 times anything is that anything (in this case X).

A lot of huff and puff and complicated looking calculations that mean nothing.

Answer 3

Any number to the 0 power is 1 so
1*((x^-2)(b^-2))^2/xb*xby
any number (^-2)^2 = that number
xb/xb*xb^2y=1/xb^2y

Answer 4

Assuming you mean

(xb^0y(x^-2b^-2)^2)/(xb(by^0) * xby)
(xy(x^(-2 * 2)b^(-2 * 2))))/(xb(b) * xby)
(xy(x^-4b^-4))/(xb^2 * xby)
(x^-4b^-4)/(xb^3)

x^(-4 - 1)b^(-4 - 3)
x^-5b^-7

ANS : 1/(x^5b^7)

Answer 5

(x*b^0*y*(x^-2*b^-2)^2)/;
(x*b(b*y^0)x*b*y)=0 assuming equal to zero
x=6
b=8
y=1

Answer 6

I think...

1 / (x * y * b^3)