Can you prove the following equation holds for all n where n is a natural number?

(2n + 1)8 + (4n + 4)9 + (6n +9)15 = (2n + 1)5 + (4n + 4)12 + (6n + 9)14

2n+1=t
4n+4=s
6n+9=k

8t+9s+15k=5t+12s+14k

3t-3s+k=0

6n+3-12n-12+6n+9=0

0=0