Had to prove something by induction. Can you please help me by explaining what magic happened after the red $\color{red}=$ in this solution?!
$$a_{n+1}=3a_n+4a_{n-1}+6=3(4^n+2(-1)^n-1)+4(4^{n-1}+2(-1)^{n-1}-1)+6\color{red}=$$
$$\color{red}=3\cdot 4^n+4^n+(8-6)(-1)^{n-1}-1=4^{n+1}+2(-1)^{n+1}-1$$
I hope that no additional info from the question is needed.
Several things happen:
The $3 \cdot 4^n$ should be clear, the $4^n$ is $4 \cdot 4^{n-1}$ from expanding the second parenthesis.
The $(8-6)(-1)^{n-1}$ arises as $4 \cdot 2 \cdot (-1)^{n-1}$ from the second parenthesis, and $3 (2 (-1)^n) = -6(-1)^{n-1}$ from the first parenthesis.
Finally the $-1$ arises as $3(-1) + 4(-1) + 6$.