SOLUTION: 7^12 - 4^12 is exactly divisible by A.33 B.34 C.35 D.36 Explain how?

Question 682117: 7^12 - 4^12 is exactly divisible by
A.33
B.34
C.35
D.36
Explain how?
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!

7^12 - 4^12

(7^6)^2 - (4^6)^2

(7^6 - 4^6)(7^6 + 4^6) ... Use the difference of squares rule

((7^3)^2 - (4^3)^2)(7^6 + 4^6)

(7^3 - 4^3)(7^3 + 4^3)(7^6 + 4^6) ... Use the difference of squares rule

(7 - 4)(7^2 + 7*4 + 4^2)(7^3 + 4^3)(7^6 + 4^6) ... Use the difference of cubes rule

(7 - 4)(49 + 28 + 16)(7^3 + 4^3)(7^6 + 4^6)

(3)(93)(7^3 + 4^3)(7^6 + 4^6)

(3)(93)(7 + 4)(7^2 - 7*4 + 4^2)(7^6 + 4^6) ... Use the sum of cubes rule

(3)(93)(7 + 4)(49 - 28 + 16)(7^6 + 4^6)

(3)(93)(11)(37)(7^6 + 4^6)

(3*11)(93)(37)(7^6 + 4^6)

(33)(93)(37)(7^6 + 4^6)

We can see that 33 is a factor of (33)(93)(37)(7^6 + 4^6), so 33 is also a factor of 7^12 - 4^12 since each line in the work shown above is equivalent to any other line (ie they represent the same number)

This means that 7^12 - 4^12 is exactly divisible by 33.

Therefore, the answer is A) 33.

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