I need to show work help

1. (k-3)(k-4) / 10(k-3)
Before simplifying, first find any numbers that k cannot be. In a fraction, the denominator cannot equal zero. To find what number makes the denominator equal zero, set the denominator equal to 0:
10(k-3) = 0
10k - 30 = 0
10k = 30
k = 3
K cannot equal three.
In the expression (k-3) is in both the numerator and denominator. (k-3) can cancel out.
This can be rewritten as (k-4) / 10 where k cannot equal 3

2. (x + 4)^2 / (x+4) (x - 2)
First, find out what values x cannot be.
(x+4)(x-2) = 0 | When you’re multiplying two things together, if one of them equals zero the end result will be zero. So you can split this into (x+4) = 0 and (x-2) = 0. x cannot equal -4 or 2
There is (x+4)^2 in the numerator and (x+4) in the denominator. (x+4)^2 = (x+4)(x+4) so only one of these (x+4) will cancel out in the numerator.
This leaves you with (x+4) / (x-2). X cannot equal -4 or 2

3. 8(x+2)^3(x-3)^3 / 4(x+2)^2(x-3)^5
First, find out want values x cannot equal
4(x+2)^2 = 0
4(x+2) = 0
(x+2) = 0
x = -2
(x-3)^5 = 0
(x-3) = 0
x = 3
X cannot equal -2 or 3

8/4 = 2
Now you have 2(x+2)^3(x-3)^3 /(x+2)^2(x-3)^5

(x+2)^3 = (x+2)(x+2)(x+2) and (x+2)^2 =(x+2)(x+2). Two of (x+2)s can cancel out, leaving:
2(x+2)(x-3)^3/((x-3)^5

(x-3)^3 = (x-3)(x-3)(x-3) and (x-3)^5 = (x-3)(x-3)(x-3)(x-3)(x-3). Three (x-3)s will cancel out, leaving you with: 2(x+2) / (x-3)^2, where x cannot equal -2 or 3

Instead of expanding the terms in parentheses squared and canceling them out, when dividing you can subtract the exponents instead (ex (x+4)^3/(x+4)^2 = x+4 because 3-2=1. (x+1)^2/(x+1)^4 = 1/(x+1)^2 because 2-4 = -2. If the subtracted number is negative the exponent belongs in the denominator)

The answer would be 27 because assuming that each package has 1 bar of soap and 1 tube of toothpaste, after 27 packages they would run out of toothpaste