We have as on RHS (a+b) we can add 4abc to the 3rd term to get (a+b) as a factor and add 2abc to 1st and 2nd term that is distributing 8abc to get
$a(b-c)^2 +b(c-a)^2+c(a-b)^2+8abc$
$=a((b-c)^2 +2bc)+b((c-a)^2+2ac)+c((a-b)^2+4ab)$
$=a((b^2+c^2)+b((c^2+a^2))+c(a+b)^2$
$=ab^2+ac^2+bc^2+ba^2+c(a+b)^2$
$=ab^2+ba^2+ac^2+bc^2+c(a+b)^2$
$=ab(a+b) + c^2(a+b) + c(a+b)^2$
$=(a+b)(ab+c^2 + c(a+b)$
$=(a+b) (c^2 + ca + bc + ab)$
$= (a+b)(b+c)(c+a)$
First let us find the number such that $\frac{n}{2}$ , $\frac{n}{3}$ ,$\frac{2n+1}{5}$ are integers
as $\frac{n}{2}$ , $\frac{n}{3}$ are integers so $\frac{n}{6}$ is integer say k
so n= 6k
As in denominator we have 2 3 and 5 so we should have mod 30
take k from 0 to 5 we get n = 12 satisfies $\frac{2n+1}{5}$ integer
so we get values n =12 , 42, 72, 102,132,162,192 .
as n is multiple of 6 $\frac{n}{2}$ , $\frac{n}{3}$ are composite we need to check $\frac{2n+1}{5}$ composite . let us compute $\frac{2n+1}{5}$
n = 12 gives 5 so no
n = 42 gives 17 so no
n = 72 gives 29 no
n = 102 gives 41 no
n = 132 gives 53 no
n = 162 gives 65 yes
n = 192 gives 77 yes
so there are 2 values
we have
$3^{(2n + 1)}+ 2^{(n + 2)}$
$=3 *3^{2n}+ 4 *2^{n}$
$=3 *3^{2n}+ 4 *2^{n}$
$=3 *9^{n}+ 4 *2^{n}$
$=3 *2^{n}+ 4 *2^{n}$ as $9 \equiv 2 \pmod 7$
$=(3 + 4) *2^{n}$
$= 7 *2^{n}$ divisible by 7
