Are you using carrots ^ to mean multiplication or do you really mean that those are all powers of powers of powers?
12a^6b^4c^2 + 15a^2b^4c^6
Rewriting that,
12(a^6)(b^4)(c^2) + 15(a^2)(b^4)(c^6)
If that is the intention, then,
We factor the numerical coefficients first.
12 and 15
There is a factor of 3 that is common to both.
So, (3 )(4 and 5)
Then the variable a.
a^6 and a^2.
There is a factor of a^2 that is common to both.
So, (a^2)(a^4 and 1).
Then the variable b.
b^4 and b^4.
b^4 is common to both.
So, (b^4)(1 and 1).
Then the variable c.
c^2 and c^6.
There is a factor of c^2 that is common to both.
So, (c^2)(1 and c^4)
Hence,
12(a^6)(b^4)(c^2) + 15(a^2)(b^4)(c^6)
= (3*a^2 *b^4 *c^2)(4*a^4*1*1 +5*1*1*c^4)
= (3a^2 b^4 c^2)(4a^4 +5c^4) ....answer.