If a, b and c are consecutive positive integers and a > b > c.

Two ways to approach here,

1) Let these 3 consecutive numbers be x-1, x and x+1 then,

(a^2 - b^2)*(b^2 - c^2)
(a-b)*(a+b)*(b-c)*(b+c)
-1*(x-1 + x)*(-1)*(x + x+1)
(2x-1)*(2x+1)
4x^2 - 1

which means adding 1 to the option and dividing by 4 should be a square, let's try the choices,

A) 22/4 => Not divisible by 4, eliminate
B) 80/4 = 20 => Not a perfect square, eliminate
C) 144/4 = 36 => Perfect square

No need to try rest of the choices.

2) Let these 3 consecutive numbers be x, x+1 and x+2 then,

(a^2 - b^2)*(b^2 - c^2)
(a-b)*(a+b)*(b-c)*(b+c)
(2x+1)*(2x+3)

If you closely look at this multiplication, it's trying to multiply 2 consecutive odd numbers like 5*7 or 13*15, so any option choice which follows similar pattern should be the correct answer.

A) 21 = 3*7 => eliminate
B) 79 => Prime number, eliminate
C) 143 = 11*13 => Correct
D) 231 = 11*21 => eliminate
E) 450 => Product of 2 odd numbers cannot be even, eliminate

IMO: C

a>b>c
a,b,c are consecutive integers, so let (a,b,c)= (n+1,n,n-1)
(a^2-b^2)*(b^2-c^2)
(n^2+1+2n-n^2)*(n^2-(n^2+1-2n))
(1+2n)*(2n-1)
((2n)^2) -1

Check the options
C) 4n^2 -1=143
4n^2=144
n^2=36
n=6

Answer: C

We are given that a, b, and c are consecutive positive integers such that a > b > c.
This means we can express them as:
a = b + 1, b = b, c = b - 1

We need to evaluate: (a^2 - b^2) * (b^2 - c^2)

Using the difference of squares identity: x^2 - y^2 = (x - y)(x + y),
we derive:
(a^2 - b^2) = (b+1)^2 - b^2 = 2b + 1
(b^2 - c^2) = b^2 - (b-1)^2 = 2b - 1

Thus, the expression simplifies to: (2b+1)(2b-1) = 4b^2 - 1

Given answer choices: 21, 79, 143, 231, 450,
we check which values satisfy 4b^2 - 1 = answer_choice.

Solving for b:
4b^2 = answer_choice + 1
b^2 = (answer_choice + 1) / 4

Checking integer values of b:

For 143:
4b^2 = 144 → b^2 = 36 → b = 6 (Valid)

Thus, the correct answer is 143.

We are given that a, b, and c are consecutive positive integers with a > b > c.
Expressing them in terms of b:
a = b + 1, b = b, c = b - 1

Evaluating (a2 - b2) * (b2 - c2) using the difference of squares:
(a2 - b2) = (b+1)2 - b2 = 2b + 1
(b2 - c2) = b2 - (b-1)2 = 2b - 1
Thus, (2b+1)(2b-1) = 4b2 - 1

Checking answer choices for 4b2 - 1:
For 143:
4b2 = 144 → b2 = 36 → b = 6 (Valid)

Answer: 143

\((a^2 - b^2)(b^2 - c^2) = (a-b)(a+b)*(b-c)(b+c)\)

From here I honestly didn't know how to solve this algebraically, so I tried to test cases starting from a=3, b=2, c=1 and adding 1 to each value.

Doing so I noticed that a-b and b-c are always 1, so we're left with

\((a+b)(b+c)\)

I observed that (a+b) and (b+c) are two odd numbers where a+b=b+c+2, so there's a difference of 2 between them.

So I listed all odd numbers with a delta of 2 starting from 3 and I multiplied each couple:

5*3= 15
7*5= 35
9*7= 63
11*9= 99
13*11= 143 => Answer C

Let (a^2 - b^2)(b^2 - c^2) be y
y =(a+b)(a-b)(b+c)(b-c)

This question seems very straight forward when assuming x-1, x, x+1. however, my brain set this up as x, x+1, x+2, which ended me with the expression 4x^2 + 8x + 3 = y. This is much more difficult to and time-consuming to solve. Am i supposed to generally assume x-1, x, x+1 for a consecutive integer question?