You should notice that the exponent 12 is a multiple of two, hence, you may write the given difference such that:

`7^12 - 4^12 = 7^(6*2) - 4^(6*2)`

Using the exponential identity `a^(b*c) = (a^b)^c` yields:

`7^12 - 4^12 = (7^6)^2 - (4^6)^2`

You should convert the difference of squares into a product such that:

`(7^6)^2 - (4^6)^2 = (7^6 - 4^6)(7^6 + 4^6)`

`7^6 - 4^6 = (7^3)^2 - (4^3)^2`

Converting the difference of squares into a product yields:

`(7^3)^2 - (4^3)^2 = (7^3 - 4^3)(7^3 + 4^3)`

You need to use the following formula, `a^3-b^3 = (a-b)(a^2+ab+b^2),` to convert the difference of cubes into a product such that:

`7^3 - 4^3 = (7-4)(49+28+16)`

`7^3 + 4^3 = (7+4)(49-28+16)`

`(7^6)^2 - (4^6)^2 = (((7-4)(49+28+16))(7^3 + 4^3))(7^6 + 4^6)`

`(7^6)^2 - (4^6)^2 = 3*93*11*37*(7^6 + 4^6)`

`7^12 - 4^12 = 33*37*93*(7^6 + 4^6)`

**Hence, evaluating the given difference of powers, you may notice that it is exactly divisible by 33, thus, you should select the answer A)33.**

If m is a even number we know that;

`a^m-b^m = (a^(m/2)-b^(m/2))(a^(m/2)+b^(m/2))`

`7^12-4^12`

`= (7^6-4^6)(7^6+4^6)`

`7^6-4^6`

`= (7^3-4^3)(7^3+4^3)`

`7^12-4^12`

`= (7^3-4^3)(7^3+4^3)(7^6+4^6)`

`a^3-b^3 = (a-b)(a^2+ab+b^2)`

`a^3+b^3 = (a+b)(a^2-ab+b^2)`

`7^3-4^3 = (7-4)(7^2+7*4+4^2) = 3*(7^2+7*4+4^2)`

`7^3+4^3 = (7+4)(7^2-7*4+4^2) = 11*(7^2-7*4+4^2)`

`7^12-4^12`

`= (7^3-4^3)(7^3+4^3)(7^6+4^6)`

`= 3*(7^2+7*4+4^2)*11*(7^2-7*4+4^2)(7^6+4^6)`

`= 33*(7^2+7*4+4^2)(7^2-7*4+4^2)(7^6+4^6)`

*So it is divisible by 33.*

## See eNotes Ad-Free

Start your **48-hour free trial** to get access to more than 30,000 additional guides and more than 350,000 Homework Help questions answered by our experts.

Already a member? Log in here.