For this exercise I will use my birth date which is January 10, 1981 (01-10-81). Therefore, I will be working with the following integers:
Let a = 1
b = -10 (negative ten)
c = 81
A) a³ - b³ = variable a and b and exponent on each of them
1³ - -10³ = for variable a I plugged in one and for variable b I plugged in a negative ten.
1 - -1000 = I then raised the integers to the given exponents which is three
1 + 1000 = subtracting a negative integer would change it into addition
1,001= bringing me to my final answer
B)
(a-b)(a² + ab + b²)
[1 - (-10)][1² + 1 (-10) + (-10)²]
[1 + 10][1 + (-10) + 100]
11 (1 - 10 + 100)
11(91)
1,001
For B) I used variables a and b. In each case for variable a I plugged in one and variable b I plugged in negative ten. After that, the two negatives turn into a positive. I added and simplified my signs, then I did the squaring and multiplying and came up with my final answer of 1,001.
C) for this expression I will use variables a, b and c. When I plug it in a equals one, b equals negative ten, and c equals eighty-one. It is a rational expression which uses a divisor of 2b - a. I then put the integers in for the variables, and evaluated the numerator and denominator, and both were negative giving me a positive answer. Which gives me an answer of 91/19 which is an improper fraction and at its lowest terms the answer is 13/3.
b - c
2b -a
-10 - 81
2 (-10) - 1
-91 13
-21 = 3 final answer (because they are both divisible by 7)
I found it interesting that the results of a³ - b³ and (a + b) (a² + ab +b²) had the same answer of 1,001, in my case. I think it is two math problems (one in addition and one in subtraction) set up differently but still produces the same answer. I think this was done to show us students that there is more than one way to set up a problem and get the same answer.