\(b,\left(x+3\right)^5=\left(x+3\right)^7\\ \Leftrightarrow\left(x+3\right)^7-\left(x+3\right)^5=0\\ \Leftrightarrow\left(x+3\right)^5\left(\left(x+3\right)^2-1\right)=0\\ \Leftrightarrow\left\{{}\begin{matrix}\left(x+3\right)^5=0\\\left(x+3\right)^2-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+3-0\\\left(x+3\right)^2=1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=-3\\x+3=\pm1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-3\\x\in\left\{-4;-2\right\}\end{matrix}\right.\)