The latex bragger is back!
From now on I am going to post the problem, solution with rendered latex, and bare solution.
We begin by seeing that both $15^6$ and $7^6$ have even powers, meaning that difference of squares is applicable.
From $a^2-b^2=(a+b)(a-b)$, we get $15^{3^{2}}-7^{3^{2}}=(15^3+7^3)(15^3-7^3)$
Another property in $a^3-b^3=(a-b)(a^2+ab+b^2)$ comes in handy to help further factorize the $(15^3-7^3)$ into $(15-7)(15^2+15\cdot 7+7^2)$
A similar property $a^3+b^3=(a + b)(a^2 - ab + b^2)$ can break apart $(15^3+7^3)$ into $(15+7)(15^2-15\cdot 7+7^2)$
Combining the two parts we end up with:
$$(15-7)(15^2+15\cdot 7+7^2)(15+7)(15^2-15\cdot 7+7^2)$$
$$(8)(225+105+49)(22)(225-105+49)$$
$$(8)(379)(22)(169)$$
Since $379$ is prime and also the largest number in the factorization, the largest prime factor of $15^6-7^6$ is $\boxed{\boxed{379}}$
Lyric(s) of the Post:I can't wait to leave this town
Cause lately I've been feeling down
The cold nights just don't feel the same
Nice factoring practice and good LaTex as usual.
ReplyDeletelol thx
Deletewizwiz quick maths
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