\((a+1)^4 - a^4\) = \([(a+1)^2 - a^2][(a+1)^2 + a^2]\) = \([(2a+1)][(a+1)^2 + a^2]\) = \([2(a+1) -1][(a+1)^2 + a^2]\)

2(a+1) -1 = one less than twice the larger integer

Answer: Option A
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I would probably opt for testing values for this question

Start with consecutive integers 1 and 2
1⁴ = 1 and 2⁴ = 16
Positive difference = 16 - 1 = 15

Now check the answer choices...

(A) one less than twice the larger integer
2(2) - 1 = 3
15 is divisible by 3
KEEP A

(B) one more than twice the larger integer
2(2) + 1 = 5
15 is divisible by 5
KEEP B

(C) one less than four times the larger integer
4(2) - 1 = 7
15 is NOT divisible by 7
ELIMINATE C

(D) one more than four times the larger integer
4(2) + 1 = 9
15 is NOT divisible by 9
ELIMINATE D

(E) one more than eight times the larger integer
8(2) + 1 = 17
15 is NOT divisible by 17
ELIMINATE E

We're left with A and B
Test another pair of consecutive integers

How about 2 and 3
2⁴ = 16 and 3⁴ = 81
Positive difference = 81 - 16 = 65

(A) one less than twice the larger integer
2(3) - 1 = 5
65 is divisible by 5
KEEP A

(B) one more than twice the larger integer
2(3) + 1 = 7
65 is NOT divisible by 7
ELIMINATE B

Answer: A

Cheers,
Brent
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\(81 - 16 = 65\)

Smaller no is 2 ; larger no is 3

(A) 65/5 = 13
(B) 65/6 = Not Divisible
(C) 65/11 = Not Divisible
(D) 65/13 = 5
(E) 65/33 = Not Divisible

Answer must be between (A) and (D) , plug in diff nos and check -

Let, Smaller no is 1 ; larger no is 2

16 - 1 = 15

(A) 15/5 = 3
(D) 15/9 = Not divisible

Hence, Answer must be (A)

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