Hello, kia!
Determine all real solutions: \(\displaystyle \;3^{2x+2}\,-\,3^{x+3}\,-\,3^x\,+\,3\;=\; 0\)
On a hunch, I tried to factor it (by "grouping") . . .
Factor: \(\displaystyle \;3^{x+2}\left(3^x\,-\,3\right)\,-\,\left(3^x\,-\,3\right)\;=\;0\)
Factor: \(\displaystyle \;\left(3^x\,-\,3\right)\left(3^{x+2}\,-\,1\right)\;=\;0\)
So we have: \(\displaystyle \,3^x\,-\,3\:=\:0\;\;\Rightarrow\;\;3^x\,=\,3\;\;\Rightarrow\;\;x\,=\,1\)
. . . . . . and: \(\displaystyle \,3^{x+2}\,-\,1\:=\:0\;\;\Rightarrow\;\;3^{x+2}\,=\,1\,=\,3^0\;\;\Rightarrow\;\;x\,+\,2\:=\:0\;\;\Rightarrow\;\;x\,=\,-2\)